蒟蒻の模板
蒟蒻 Rainbow-Automaton 的模板
\(\text{2023-10}\) 备战 \(\text{csp-s}\)
只是目前会的
然而目前啥也不会...
代码注意事项
-
不要使用
using namespace std; -
min和max都可以直接std::minstd::max吧 -
关同步
#define fastread std::ios_sync_with_stdio (false); std::cin.tie (nullptr);upd: 更为简洁的写法是
std::cin.tie(nullptr) -> sync_with_stdio(false); -
学习 jiangly
using i64 = long long
杂项
对拍
inline bool pai () {
// Generate
system ("gen.exe > test.in");
// display ();
// Run
system ("test_1.exe < test.in > test_1.out");
system ("test_2.exe < test.in > test_2.out");
if (system ("fc test_1.out test_2.out")) {
system ("pause");
return false;
}
return true;
}
毫秒级随机数
通常作为对拍时的随机数生成器
std::mt19937 rng(std::chrono::steady_clock::now ().time_since_epoch ().count ());
inline int getRand (int l, int r) {
return std::uniform_int_distribution<int> (l, r) (rng);
}
计时
防止 TLE
inline void getTime () {
system ("gen.exe > test.in");
auto s = std::chrono::steady_clock::now ();
system ("test_2.exe < test.in > test_2.out");
auto t = std::chrono::steady_clock::now ();
auto duration = std::chrono::duration_cast <std::chrono::microseconds> (t - s);
double cost = (double) duration.count () / 1000;
std::cout << cost << "\n";
}
图论
存图
这种东西真的需要记吗
写一个神秘的封装好的图
template<int maxn, int maxm>
struct Graph {
struct Edge {
int u, v;
int pre;
} es[maxm << 1];
int last[maxn], cnt;
Graph () { cnt = 0; memset (last, 0, sizeof (last)); }
inline void addEdge (int u, int v) {
es[++cnt] = Edge { u, v, last[u] };
last[u] = cnt;
}
};
然而我大部分时侯都不会用到封装好的图...
一般情况下只会直接写 Graph 里面的东西
最短路
Dijkstra
struct Node {
int id;
int dis;
friend bool operator < (Node a, Node b) {
return a.dis > b.dis; // 通过改符号的方向实现小根堆...
}
};
std::priority_queue<Node> q;
int dis[maxn];
bool vis[maxn];
inline void dij (int S) {
memset (vis, 0, sizeof (vis));
memset (dis, 0x3f, sizeof (dis)); dis[S] = 0;
q.push (Node { S, dis[S] });
while (not q.empty()) {
int now = q.top ().id; q.pop ();
if (vis[now]) { continue; }
vis[now] = true;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (dis[t] > dis[now] + es[i].w) {
dis[t] = dis[now] + es[i].w;
q.push (Node { t, dis[t] });
}
}
}
}
某个广为人所知的最短路算法
判负环
不判负环最好别用
才发现我单源最短路都是用 Dijkstra 过的
甚至找不到一个SPFA的板子
std::queue<int> q;
bool inq[maxn];
int times[maxn];
int dis[maxn];
inline bool spfa (int S) {
q.push (S);
memset (inq, 0, sizeof (inq)); inq[S] = true;
memset (times, 0, sizeof (times)); times[S] = 0;
for (int i = 1; i <= n + 1; i++) { if (i != S) { dis[i] = inf; } }
while (not q.empty ()) {
int now = q.front (); q.pop ();
inq[now] = false;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (dis[t] > dis[now] + es[i].w) {
dis[t] = dis[now] + es[i].w;
if (not inq[t]) {
inq[t] = true;
times[t] ++; if (times[t] > n + 1) { return false; } // 此处n + 1是因为多了一个超级源
q.push (t);
}
}
}
}
return true;
}
差分约束
也许是唯一能用上SPFA的地方吧...
可以解出 \(m\) 个形如 \(X_u - X_v \leq Y\) 的不等式组成的不等式组的一个解
自然也可以判断是否有解
实质就是 \(u - v \leq w \implies u \leq v + w \implies Edge_{v, u, w}\)
\(v\) 到 \(u\) 连一条长度为 \(w\) 的边
inline void diff_constraints () { // 差分约束
std::cin >> n >> m;
for (int i = 1; i <= m; i++) {
int u, v, w; std::cin >> u >> v >> w;
addEdge (v, u, w);
}
// 以n + 1作为超级源点
for (int i = 1; i <= n; i++) { addEdge (n + 1, i, 0); }
if (not spfa (n + 1)) { std::cout << "NO\n"; }
else {
for (int i = 1; i <= n; i++) { std::cout << dis[i] << " "; }
}
}
Floyd
int dis[maxn][maxn];
inline void init () {
memset (dis, 0x3f, sizeof (dis));
for (int i = 1; i <= n; i++) { dis[i][i] = 0; }
}
inline void floyd () {
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
dis[i][j] = std::min (dis[i][j], dis[i][k] + dis[k][j]);
}
}
}
}
注意先 init() 再加边
倍增Floyd
实际上是矩阵快速幂
把 \(\text{Floyd}\) 原来的用 \(dp[k][i][j]\) 表示前 \(k\) 个点中 \(i\) 到 \(j\) 的最短路
改为用 \(dp[s][i][j]\) 表示长度为 \(s\) 的 \(i\) 到 \(j\) 的最短路
显然
\(dp[s][i][j]\) 可以由 \(\min\limits_{k = 1}^{n} \{dp[s - 1][i][k] + dp[1][k][j]\}\) 转移过来
然后我们发现转移的过程和矩阵乘法特别像
用快速幂求出 \(dp^k\) 就可以得到长为 \(k\) 的路径条数
传递闭包
实际上就是 \(01\) 矩阵上的 \(\text{Floyd}\)
只是求最短路改成求连通性了
std::bitset<maxn> a[maxn];
inline void floydClosure () {
for (int j = 1; j <= n; j++) {
for (int i = 1; i <= n; i++) {
if (a[i][j]) { // i 能到 j
a[i] |= a[j]; // 用j更新i
}
}
}
}
上面的代码是用
bitset写的, 与传统 \(\text{Floyd}\) 略有不同但是其基本思想是类似的
生成树
Kruskal
struct Edge {
int u, v, w;
friend bool operator < (Edge a, Edge b) {
return a.w < b.w;
}
};
std::vector<Edge> es;
inline int kruskal () {
std::sort (es.begin (), es.end ());
init ();
int ans = 0;
int tot = 0;
for (auto e : es) {
int u = find (e.u);
int v = find (e.v);
if (u == v) { continue; }
fa[u] = v;
ans += e.w;
tot ++;
if (tot == n - 1) { return ans; }
}
return -1;
}
其中
init()就是并查集的init()并查集的写法看数据结构部分吧
欧拉回路
一定要加当前弧优化!
否则复杂度其实是假的
依然记得在mx想出正解但是欧拉回路写挂了
std::vector<int> g[maxn];
int ind[maxn], otd[maxn];
int st[maxn];
std::vector<int> euler;
void dfs (int now, int eid) {
for (int i = st[now]; i < g[now].size (); i = st[now]) { // g[now] 中编号一定是单增的
st[now] = i + 1;
int t = g[now][i];
dfs (es[t].v, t);
}
euler.push_back (eid);
}
upd: 换一个更人性的写法
std::vector <int> g[maxn];
int ind[maxn], otd[maxn];
std::vector <int> euler;
void dfs (int now) {
db (now);
while (not g[now].empty ()) {
int t = g[now].back ();
g[now].pop_back (); // 自带当前弧优化
dfs (t);
}
euler.push_back (now);
}
upd: 判断
-
有向图欧拉路径的判断
bool check () { int dcnt1 = 0; int dcnt2 = 0; for (int i = 1; i <= tot; i++) { if (ind[i] - otd[i] == 0) { continue; } if (ind[i] - otd[i] == 1) { dcnt1 ++; continue; } if (ind[i] - otd[i] == -1) { dcnt2 ++; start = i; continue; } return false; } if (dcnt1 > 1 or dcnt2 > 1 or dcnt1 + dcnt2 == 1) { return false; } } -
判断是否存在无向图欧拉回路 : 是否所有点入度等于出度
最后记得判断求出来的长度是否合法
Tarjan 系列
强连通分量
对于一个有向图, 一个强连通分量 \(S\) 是指极大的保证 \(\forall u, v \in S\) 存在 \(u \to v\) 的路径 和 \(v \to u\) 的路径的点集
int low[maxn], dfn[maxn], dpos;
int stk[maxn], spos;
bool ins[maxn];
int belong[maxn], scc_cnt;
std::set<int> scc[maxn];
void tarjan (int now) {
low[now] = dfn[now] = ++dpos;
stk[++spos] = now;
ins[now] = true;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (not dfn[t]) {
tarjan (t);
low[now] = std::min (low[now], low[t]);
} else if (ins[t]) {
low[now] = std::min (low[now], dfn[t]);
}
}
if (low[now] == dfn[now]) {
scc_cnt ++;
do {
belong[now] = scc_cnt;
scc[scc_cnt].insert (now);
ins[now] = false;
now = stk[spos--];
} while (low[now] != dfn[now]);
}
}
2-SAT
用 \(1\) 到 \(n\) 表示 \(a_i\) 取 \(0\)
用 \(n + 1\) 到 \(2 \times n\) 表示 \(a_{i - n}\) 取 \(1\)
for (int i = 1; i <= 2 * n; i++) {
if (not dfn[i]) { tarjan (i); }
}
if (not check ()) { std::cout << "IMPOSSIBLE" << "\n"; return 0; }
std::cout << "POSSIBLE" << "\n";
for (int i = 1; i <= n; i++) {
if (scc[i + n] < scc[i]) { std::cout << 1 << " "; }
else { std::cout << 0 << " "; }
}
割点
int low[maxn], dfn[maxn], dpos;
bool cut[maxn];
void tarjan (int now, int fa) {
low[now] = dfn[now] = ++dpos;
int sons = 0;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (not dfn[t]) {
tarjan (t, now);
low[now] = std::min (low[now], low[t]);
sons ++;
if (low[t] >= dfn[now]) { cut[now] = true; }
} else { // 此处可以考虑父亲
low[now] = std::min (low[now], dfn[t]);
}
}
if (fa == 0 and sons < 2) { cut[now] = false; }
}
点双连通分量
int low[maxn], dfn[maxn], dpos;
int stk[maxn], spos;
int belong[maxn], vdcc_cnt;
std::vector<int> vdcc[maxn];
void tarjan (int now, int fa) {
low[now] = dfn[now] = ++dpos;
stk[++spos] = now;
int sons = 0;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (not dfn[t]) {
tarjan (t, now);
low[now] = std::min (low[now], low[t]);
sons ++;
if (low[t] >= dfn[now]) {
++vdcc_cnt;
int last = 0;
while (last != t) {
int top = stk[spos--];
belong[top] = vdcc_cnt;
vdcc[vdcc_cnt].push_back (top);
last = top;
}
vdcc[vdcc_cnt].push_back (now);
}
} else {
low[now] = std::min (low[now], dfn[t]);
}
}
if (fa == 0 and sons == 0) { vdcc_cnt ++; vdcc[vdcc_cnt].push_back (now); }
}
桥
int low[maxn], dfn[maxn], dpos;
bool bridge[maxm << 1];
void tarjan (int now, int in_edge) {
low[now] = dfn[now] = ++dpos;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (not dfn[t]) {
tarjan (t, i);
low[now] = std::min (low[now], low[t]);
if (low[t] > dfn[now]) { bridge[i] = bridge[i ^ 1] = true; }
} else if (i != (in_edge ^ 1)) {
low[now] = std::min (low[now], dfn[t]);
}
}
}
inline void init () { cnt = 1; }
因为涉及到快速找反向边, 一定要把
cnt初始化为1!
边双连通分量
int low[maxn], dfn[maxn], dpos;
bool bridge[maxm << 1];
void tarjan (int now, int in_edge) {
low[now] = dfn[now] = ++dpos;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (not dfn[t]) {
tarjan (t, i);
low[now] = std::min (low[now], low[t]);
if (low[t] > dfn[now]) { bridge[i] = bridge[i ^ 1] = true; }
} else if (i != (in_edge ^ 1)) {
low[now] = std::min (low[now], dfn[t]);
}
}
}
int belong[maxn], edcc_cnt;
std::vector<int> edcc[maxn];
void dfs (int now) {
belong[now] = edcc_cnt;
edcc[edcc_cnt].push_back (now);
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (bridge[i]) { continue; }
if (belong[t]) { continue; }
dfs (t);
}
}
inline void init () { cnt = 1; }
网络流
网络流的加边方式与普通图论问题的加边方式略有不同
struct Edge { int u, v; int pre; long long flow; } es[maxm << 1]; int last[maxn], cur[maxn], cnt = 1; int S, T; inline void _addEdge (int u, int v, long long cap) { es[++cnt] = Edge { u, v, last[u], cap }; last[u] = cnt; } inline void addEdge (int u, int v, long long cap) { _addEdge (u, v, cap); _addEdge (v, u, 0); }
Dinic
int dep[maxn];
bool bfs () {
std::queue<int> q; q.push (S);
memset (dep, 0x3f, sizeof (dep)); dep[S] = 1;
while (not q.empty ()) {
int now = q.front (); q.pop ();
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (not es[i].flow) { continue; }
if (dep[t] != 0x3f3f3f3f) { continue; }
dep[t] = dep[now] + 1;
q.push (t);
}
}
return dep[T] != 0x3f3f3f3f;
}
long long dfs (int now, long long now_flow) {
if (now == T or not now_flow) { return now_flow; }
long long res = 0;
for (int &i = cur[now]; i; i = es[i].pre) {
int t = es[i].v;
if (not es[i].flow) { continue; }
if (dep[t] != dep[now] + 1) { continue; }
long long t_flow = dfs (t, std::min (now_flow, es[i].flow));
if (t_flow) {
es[i].flow -= t_flow;
es[i ^ 1].flow += t_flow;
now_flow -= t_flow;
res += t_flow;
if (not now_flow) { return res; }
}
}
return res;
}
inline long long Dinic (int _s, int _t) {
S = _s, T = _t;
long long res = 0;
while (bfs ()) {
memcpy (cur, last, sizeof (last));
res += dfs (S, 0x3f3f3f3f);
}
return res;
}
网络单纯形
把之前写的代码粘过来了
码风可能有点不太一样
而且好像使用 \(\text{LCT}\) 实现会更容易一些...
然而我学网络单纯形的时候并不会lct
const long long inf = 1e9;
namespace NetworkSimplex {
struct Edge {
int u, v;
int pre;
int flow;
int cost;
} es[maxm];
int last[maxn], cnt = 1;
inline void _addEdge (int u, int v, int cap, int cost) {
es[++cnt] = Edge { u, v, last[u], cap, cost };
last[u] = cnt;
}
inline void addEdge (int u, int v, int cap, int cost) {
_addEdge (u, v, cap, cost);
_addEdge (v, u, 0, -cost);
}
int fa[maxn], faEdge[maxn]; // fa[u] : u的父亲 | faEdge[u] : 父亲指向u的边
int mark[maxn], curTime = 1; // mark[u] : 节点u的时间戳 | curTime : 当前时间
// 先找出基解 (随便找一棵生成树)
// 此时mark的作用是标记是否访问过
void initTree (int now) {
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (not es[i].flow) { continue; }
if (mark[t]) { continue; }
mark[t] = 1;
fa[t] = now;
faEdge[t] = i;
initTree (t);
}
}
int piCache[maxn]; // 缓存一下pi优化
int pi (int now) { // 当前节点到根节点的代价和
if (mark[now] == curTime) { return piCache[now]; } // 当前节点的时间戳是当前时间 (piCache里的东西没有过时), 直接返回
mark[now] = curTime; // 当前点经过更新已经是最新的了
return piCache[now] = es[faEdge[now]].cost + pi(fa[now]); // 当前点的pi是从根节点到父亲的代价 + 父亲到当前点的代价
}
inline long long pushFlow (int eid) { // 沿着给定非树边的编号eid形成的环推流
++curTime; // 时间戳增加
// 找根
int root = es[eid].u;
while (root) { mark[root] = curTime; root = fa[root]; } // 向上跳并在跳的路径上打标记
// 找lca
int lca = es[eid].v;
while (mark[lca] != curTime) { mark[lca] = curTime; lca = fa[lca]; } // 向上跳, 一直跳到被标记过 (也就是说要跳到在u到根的路径上),, 显然第一个跳到的点是lca
// 找出基的边 (流量最小的边)
int minFlow = es[eid].flow; // 能流的流量 (最小的流量)
int delNode = 0; // 被删去的边的对应点
int flag = 2; // flag = 2 : 没找到 (当前边就是) | 0 : 在lca到u的路径上 | 1 : 在v到lca的路径上
std::vector <int> circle; // 存环上的边
// 从u向上找lca往u方向的流量最小的边
for (int now = es[eid].u; now != lca; now = fa[now]) {
int nowEdge = faEdge[now]; // 因为是lca往u方向流所以是u的父亲到u的边
circle.push_back (nowEdge);
if (minFlow > es[nowEdge].flow) {
minFlow = es[nowEdge].flow;
flag = 0;
delNode = now;
}
}
// 从v想上找v往lca方向的流量最小的边
for (int now = es[eid].v; now != lca; now = fa[now]) {
int nowEdge = faEdge[now] ^ 1; // 因为是v往lca方向流所以是v到父亲的边
circle.push_back (nowEdge);
if (minFlow > es[nowEdge].flow) {
minFlow = es[nowEdge].flow;
flag = 1;
delNode = now;
}
}
circle.push_back (eid);
// 给环上每个边增加流量并统计费用
long long cost = 0;
for (auto now : circle) {
es[now].flow -= minFlow;
es[now ^ 1].flow += minFlow;
cost += minFlow * es[now].cost;
}
if (flag == 2) { return cost; } // 最小边是当前边, 不许要改变树的结构
// 改变树的结构 (加入入基边, 删除出基边)
// 实际上就是翻转树的结构 (此处我们直接暴力翻转)
// 如果最小边是在lca到u的路径上, 我们实际上要对v->u->delNode的路径进行更新;
// 如果最小边是在v到lca的路径上, 我们实际上要对u->v->delNode的路径进行更新
// 这里有一个trick
// 我们发现, 在第一种情况里的u->delNode和第二种情况里v->delNode的路径中, 我们原有的faEdge[x]都是从我们要跳到的点, 指向当前点x
// 如果我们能统一边的方向(从我们要跳到的点指向当前点), 翻转整条链的操作就会变得非常好处理
// 然后我们又发现
// 第一种情况里u->v的这条边恰好满足从我们要跳到的点u到我们当前的点v这一条件
// 第二种情况正好相反
// 所以第二种情况下, 我们先对我们要加入的这条边u->v翻转一下, 取他的反向边v->u
// 这样就可以统一处理了
eid = eid ^ flag; // 所以我们直接异或上flag (如果在lca到u的路径上, flag就等于0, 异或上0等于没有异或)
int u = es[eid].u;
int v = es[eid].v;
int lastNode = v; // 上一个操作的点
int lastEdge = eid; // 上一个边
int nextNode = 0; // 下一个点
for (int now = u; lastNode != delNode; now = nextNode) {
nextNode = fa[now];
mark[now] --; // 信息过期, 时间戳--
lastEdge ^= 1; // 将上一条边翻转
swap (lastEdge, faEdge[now]);
fa[now] = lastNode;
lastNode = now;
}
return cost;
}
struct MCMF {
int minCost, maxFlow;
};
inline MCMF simplex (int S, int T) {
addEdge (T, S, inf, -inf); // 加这条辅助边保证最大流
initTree (T); // 寻找基解
mark[T] = ++curTime;
fa[T] = 0; // 钦定T是根
long long cost = 0; // 由于一开始加了一条费用为-inf的边, 费用可能很大, 最好开long long
while (true) {
bool flag = false;
for (int i = 2; i <= cnt; i++) {
if (es[i].flow and es[i].cost + pi(es[i].u) - pi(es[i].v) < 0) { // 寻找可以形成负环的边
flag = true;
cost += pushFlow (i);
}
}
if (not flag) { break; } // 没有找到可以形成负环推流的边, 停止
}
// 最后加上的边反向边 (S -> T) 的flow就是我们的最大流
// (因为我们这条边加的意义是使得任意S到T的流都可以与它形成负环, 所以每次推流的时候都会把增加的流量加到它上面去)
int maxFlow = es[cnt].flow;
int minCost = int ((long long) es[cnt].flow * inf + cost); // 需要加上 es[cnt].flow * inf 抵消掉一开始那条边的费用
return MCMF { minCost, maxFlow };
}
inline void clear () {
memset (mark, 0, sizeof(mark)); curTime = 1;
memset (piCache, 0, sizeof(piCache));
memset (last, 0, sizeof(last)); cnt = 1;
}
}
字符串
啥也不会
前缀函数
inline std::vector<int> get_pi(const std::string &s) {
std::vector<int> pi(s.length());
rep (i, 2, (int) s.length() - 1) {
int j = pi[i - 1];
while (j and s[j + 1] != s[i]) j = pi[j];
if (s[j + 1] == s[i]) j++;
pi[i] = j;
}
return pi;
}
AC 自动机
就是前缀自动机
template <int maxn>
struct ACAM {
struct Node {
int ch[26];
int fail;
} tr[maxn];
int rt, tot;
inline void init() { rt = tot = 1; std::memset(tr, 0, sizeof tr); }
ACAM() { init(); }
inline int ins(const std::string &s) {
int now = rt;
for (auto c : s) {
int &t = tr[now].ch[c - 'a'];
if (not t) t = ++tot;
now = t;
}
return now;
}
inline void build() {
for (auto &t : tr[0].ch) t = rt;
std::queue<int> q;
q.push(rt);
while (not q.empty()) {
int now = q.front(); q.pop();
for (int i = 0; i < 26; i++) {
int &t = tr[now].ch[i];
int f = tr[tr[now].fail].ch[i];
if (not t) t = f;
else {
tr[t].fail = f;
q.push(t);
}
}
}
}
};
后缀自动机
struct SAM {
struct Node {
int ch[26];
int fa;
int len;
} ns[maxn];
int last, root, tot;
// 附加信息
i64 cnt[maxn];
SAM () { last = root = tot = 1; memset (cnt, 0, sizeof (cnt)); }
inline void add (int c) {
int p = last;
int np = last = ++tot;
ns[np].len = ns[p].len + 1;
cnt[np] = 1;
for (; p and not ns[p].ch[c]; p = ns[p].fa) { ns[p].ch[c] = np; }
if (not p) { ns[np].fa = root; }
else {
int q = ns[p].ch[c];
if (ns[q].len == ns[p].len + 1) { ns[np].fa = q; }
else {
int nq = ++tot;
ns[nq] = ns[q]; ns[nq].len = ns[p].len + 1;
ns[q].fa = ns[np].fa = nq;
for (; p and ns[p].ch[c] == q; p = ns[p].fa) { ns[p].ch[c] = nq; }
}
}
}
inline void getParentTree (Graph& tr) {
for (int i = 2; i <= tot; i++) { tr.addEdge (ns[i].fa, i); }
}
};
回文自动机
namespace Palindrome_Automaton {
struct Node {
int len, cnt;
int next[26];
int fail;
};
Node pam[maxn];
int cnt, last;
inline void init () {
cnt = 1, last = 0; // 奇根的前一个是偶根
pam[0].len = 0;
pam[1].len = -1; // 两边同时加入一个
pam[0].fail = 1; // 奇根不会有fail
}
// decrypt (x) 只是为了满足模板题里面的强制在线而已
#define decrypt(x) (s[i] + pam[last].cnt - 97) % 26 + 97
#define get_fail(x) while (s[i] != s[i - pam[x].len - 1]) { x = pam[x].fail; }
inline std::vector<int> solve (string s) {
std::vector<int> res;
for (int i = 0; i < s.size(); i++) {
if (i >= 1) { s[i] = decrypt(i); }
int a = last; get_fail(a);
while (b) {
if (b & 1) { (res *= base) %= mod; }
(base *= base) %= mod;
while (b) {
if (b & 1) { (res *= base) %= mod; }
(base *= base) %= mod;
while (b) {
if (b & 1) { (res *= base) %= mod; }
(base *= base) %= mod;
while(cnt<n/g){
int tcnt=(n-i-1)/k+1;
int l=1,r=tcnt+1;
while(l<r){
int mid=(l+r)/2;
if(i+(mid-1)*k>=cnt+mid) r=mid;
else l=mid+1;
}
if(l<=tcnt){
int st=i+(l-1)*k,ed=i+(tcnt-1)*k;
ans+=1LL*(st+ed)*(tcnt-l+1)/2+(tcnt-l+1);
}
cnt+=tcnt;
i=(i+(tcnt-1)*k+k)%n;
}
return ans;
}
int main(){
cin.tie(nullptr)->sync_with_stdio(false);
int T;
cin>>T;
while(cnt<n/g){
int tcnt=(n-i-1)/k+1;
int l=1,r=tcnt+1;
while(l<r){
int mid=(l+r)/2;
if(i+(mid-1)*k>=cnt+mid) r=mid;
else l=mid+1;
}
if(l<=tcnt){
int st=i+(l-1)*k,ed=i+(tcnt-1)*k;
ans+=1LL*(st+ed)*(tcnt-l+1)/2+(tcnt-l+1);
}
cnt+=tcnt;
i=(i+(tcnt-1)*k+k)%n;
}
return ans;
}
int main(){
cin.tie(nullptr)->sync_with_stdio(false);
int T;
cin>>T;
if (not pam[a].next[s[i] - 'a']) { // 没有就新建节点
int b = pam[a].fail; get_fail(b);
cnt ++;
pam[cnt].fail = pam[b].next[s[i] - 'a'];
pam[cnt].cnt = pam[pam[cnt].fail].cnt + 1;
pam[cnt].len = pam[a].len + 2;
pam[a].next[s[i] - 'a'] = cnt;
}
last = pam[a].next[s[i] - 'a'];
res.push_back(pam[last].cnt);
}
return res;
}
}
数学
快速幂
i64 ksm (i64 a, i64 b, i64 mod) {
i64 res = 1;
i64 base = a;
while (b) {
if (b & 1) { (res *= base) %= mod; }
(base *= base) %= mod;
b >>= 1;
}
return res;
}
欧拉函数
i64 phi (i64 n) {
i64 ans = n;
for (i64 p = 2; p * p <= n; p++) {
if (n % p != 0) { continue; }
ans = ans / p * (p - 1);
while (n % p == 0) { n /= p; }
}
if (n > 1) { ans = ans / n * (n - 1); }
return ans;
}
逆元
这里使用欧拉函数求逆元
不会写exgcd导致的
你说的对, 但是 \(a ^ {\varphi(b)} = 1 \pmod b\)
所以我们可以直接求逆元
i64 inv (i64 a, i64 b) {
return Ksm::ksm (a, phi (b) - 1, b);
}
exgcd
i64 exgcd (i64 a, i64 b, i64 &x, i64 &y) {
if (b == 0) { x = 1, y = 0; return a; }
i64 d = exgcd (b, a % b, y, x);
y -= x * (a / b);
return d;
}
于是, 上面的逆元也可以用exgcd求
因为
\[a \times inv_a \equiv 1 \pmod b \\
a \times inv_a + b \times y = 0
\]
所以我们很容易用exgcd求出 \(inv_a\)
inline i64 inv (i64 a, i64 b) {
i64 x, y; exgcd (a, b, x, y);
return (x + b) % b;
}
线性筛
std::bitset<maxn> not_prime;
std::vector<int> primes;
inline void get_primes () {
not_prime.reset ();
not_prime[1] = true;
for (int i = 2; i <= n; i++) {
if (not not_prime[i]) { primes.push_back (i); }
for (auto p : primes) {
if (i * p > n) { break; }
not_prime[i * p] = true;
if (i % p == 0) { break; }
}
}
}
筛莫比乌斯函数
用线性筛筛出莫比乌斯函数
顺便求出前缀和
inline void get_mu () {
not_prime.reset ();
int n = maxn - 5;
mu[1] = 1;
not_prime[1] = true;
for (int i = 2; i <= n; i++) {
if (not not_prime[i]) { primes.push_back (i); mu[i] = -1; }
for (auto p : primes) {
if (i * p > n) { break; }
not_prime[i * p] = true;
mu[i * p] = -1 * mu[i];
if (i % p == 0) { mu[i * p] = 0; break; }
}
}
for (int i = 1; i <= n; i++) { sum[i] = sum[i - 1] + mu[i]; }
}
中国剩余定理
只会大数翻倍法
所以这其实是假的中国剩余定理
orz钟神太强了%%%
inline bool merge (i64 &m1, i64 &a1, i64 m2, i64 a2) {
if (m1 < m2) { std::swap (m1, m2); std::swap (a1, a2); }
while (a1 % m2 != a2) { a1 += m1; }
m1 = lcm (m1, m2);
return true;
}
真的超级好写啊啊啊
upd: 判无解版本:
inline bool merge (i64& m1, i64& a1, i64 m2, i64 a2) {
if (m1 < m2) { std::swap (m1, m2); std::swap (a1, a2); }
int tot = 0;
while (a1 % m2 != a2) {
if (++tot > m2 + 1) { return false; }
a1 += m1;
}
m1 = lcm (m1, m2);
return true;
}
扩展中国剩余定理
不用害怕, zhx的大数翻倍法本身就能合并两个同余方程
所以合并部分代码跟上面一样
高维前缀和
for (int i = 0; i < 22; i++) {
for (int S = 0; S < (1 << 22); S++) {
if (S & (1 << i)) { sum[S] = sum[S] + sum[S ^ (1 << i)]; }
}
}
异或线性基
struct Basis {
u64 b[63];
inline bool ins(u64 x) {
per (i, 60, 0) {
if (x & (1ull << i)) {
if (not b[i]) return (b[i] = x), true;
else x ^= b[i];
}
}
return false;
}
inline u64 mx(u64 x) {
per (i, 60, 0) x = std::max(x, x ^ b[i]);
return x;
}
} basis;
高斯消元
f64 a[maxn][maxn];
const f64 eps = 1e-10;
inline int gauss() {
int rnkA = 0; // 系数矩阵的秩
rep (i, 1, n) {
int t = rnkA + 1;
rep (j, rnkA + 1, n) {
if (std::fabs(a[j][i]) > std::fabs(a[t][i])) t = j;
}
if (std::fabs(a[t][i]) < eps) continue;
rnkA++;
std::swap(a[rnkA], a[t]);
per (j, n + 1, i) a[rnkA][j] /= a[rnkA][i];
rep (j, rnkA + 1, n) {
per (k, n + 1, i) a[j][k] -= a[rnkA][k] * a[j][i];
}
}
if (rnkA == n) {
per (i, n - 1, 1) rep (j, i + 1, n) a[i][n + 1] -= a[i][j] * a[j][n + 1];
return -1; // 有唯一解
} else {
rep (i, rnkA + 1, n) if (std::fabs(a[i][n + 1]) > eps) return 0; // 无解
return 1; // 无穷多解
}
}
多项式
NTT
namespace NTT {
const i64 g = 3, mod = 998244353;
const int maxb = 21;
const int maxn = 1 << maxb;
i64 cache[maxn << 1], *pw = cache + maxn;
inline void init() {
i64 v = ksm(g, (mod - 1) / maxn);
pw[0] = pw[0 - maxn] = 1;
rep (i, 1, maxn - 1) pw[i] = pw[i - maxn] = pw[i - 1] * v % mod;
}
inline void DFT(std::vector<i64> &a, int c) {
int len = a.size();
std::vector<i64> org = a;
rep (i, 0, len - 1) {
int pos = 0;
rep (b, 0, maxb - 1) if (i & (1 << b)) pos += (len >> (b + 1));
a[pos] = org[i];
}
for (int h = 2; h <= len; h <<= 1) {
for (int i = 0; i < len; i += h) {
for (int j = i; j < i + (h >> 1); j++) {
i64 u = a[j] % mod;
i64 v = a[j + (h >> 1)] % mod * pw[(maxn / h) * c * (j - i)] % mod;
a[j] = (u + v >= mod ? u + v - mod : u + v);
a[j + (h >> 1)] = (u - v < 0 ? u - v + mod : u - v);
}
}
}
if (c == -1) {
i64 inv_n = inv(len);
rep (i, 0, len - 1) (a[i] *= inv_n) %= mod;
}
}
inline std::vector<i64> mul(std::vector<i64> a, std::vector<i64> b) {
int len = 1;
while (len < a.size() + b.size()) len <<= 1;
a.resize(len); b.resize(len);
DFT(a, 1);
DFT(b, 1);
std::vector<i64> c(len);
rep (i, 0, len - 1) c[i] = a[i] * b[i] % mod;
DFT(c, -1);
return c;
}
}
数据结构
并查集
template <int maxn>
struct UnionFindSet {
int fa[maxn];
inline void init () {
for (int i = 1; i <= n; i++) { fa[i] = i; }
}
int find (int x) {
if (fa[x] == x) { return fa[x]; }
else { return fa[x] = find (fa[x]); }
}
inline void merge (int x, int y) {
fa[find (y)] = find (x);
}
inline bool same (int x, int y) {
return find (x) == find(y);
}
};
树状数组
template <int maxsiz>
struct BIT {
int tr[maxsiz];
inline int lowbit (int x) { return x & (-x); }
inline void add (int pos, int val) { for (int i = pos; i <= n; i += lowbit(i)) { tr[i] += val; } }
inline int _query (int pos) { int res = 0;for (int i = pos; i; i -= lowbit(i)) { res += tr[i]; } return res; }
inline int query (int l, int r) { return _query (r) - _query (l - 1); }
};
上面只给出了单点加区间求和的版本
实际上, 树状数组还可以通过神秘的技术实现区间加单点查和 区间加区间求和
甚至可以通过跳
lowbit实现 区间最值查询然而实现十分复杂, 失去了树状数组本身简洁的优势
而且复杂度好像也不是很对(曾被hzhl狠狠质疑)所以不如写线段树 👇
线段树
template <int maxn, typename val_t>
struct SegmentTree {
struct Node {
val_t sum;
val_t tag; bool cov;
} tr[maxn << 2];
inline void pushUp (int now) {
tr[now].sum = tr[now << 1].sum + tr[now << 1 | 1].sum;
}
inline void update (int now, int l, int r, val_t val) {
tr[now].sum += val * (r - l + 1);
tr[now].tag += val;
tr[now].cov = true;
}
inline void pushDown (int now, int l, int r) {
if (not tr[now].cov) { return; }
int mid = (l + r) >> 1;
update (now << 1, l, mid, tr[now].tag);
update (now << 1 | 1, mid + 1, r, tr[now].tag);
tr[now].tag = 0; tr[now].cov = false;
}
void build (int now, int l, int r) {
if (l == r) {
tr[now] = Node { a[l], 0, false };
return;
}
int mid = (l + r) >> 1;
build (now << 1, l, mid);
build (now << 1 | 1, mid + 1, r);
pushUp (now);
}
void modify (int now, int l, int r, int L, int R, val_t val) {
if (L <= l and r <= R) { update (now, l, r, val); return; }
pushDown (now, l, r);
int mid = (l + r) >> 1;
if (L <= mid) { modify (now << 1, l, mid, L, R, val); }
if (R > mid) { modify (now << 1 | 1, mid + 1, r, L, R, val); }
pushUp (now);
}
val_t query (int now, int l, int r, int L, int R) {
if (L <= l and r <= R) { return tr[now].sum; }
pushDown (now, l, r);
int mid = (l + r) >> 1;
val_t res = 0;
if (L <= mid) { res += query (now << 1, l, mid, L, R); }
if (R > mid) { res += query (now << 1 | 1, mid + 1, r, L, R); }
return res;
}
};
使用的时候, 只需要
SegmentTree<maxn, i64> tr就行
权值线段树
其实是动态开点线段树
template <int maxn, typename val_t>
struct SegmentTree {
struct Node {
int lson, rson;
val_t sum;
} tr[(maxn << 4) + 5];
int root;
int tot;
SegmentTree () { tot = 0; }
inline void pushUp (int now) { tr[now].sum = tr[tr[now].lson].sum + tr[tr[now].rson].sum; }
void modify (int &now, val_t l, val_t r, val_t pos, val_t val) {
if (not now) { now = ++tot; }
if (l == r) { tr[now].sum += val; return; }
val_t mid = (l + r) >> 1;
if (pos <= mid) { modify (tr[now].lson, l, mid, pos, val); }
if (pos > mid) { modify (tr[now].rson, mid + 1, r, pos, val); }
pushUp (now);
}
val_t query (int now, val_t l, val_t r, val_t L, val_t R) {
if (not now) { return 0; }
if (L <= l and r <= R) { return tr[now].sum; }
val_t mid = (l + r) >> 1;
val_t res = 0;
if (L <= mid) { res += query (tr[now].lson, l, mid, L, R); }
if (R > mid) { res += query (tr[now].rson, mid + 1, r, L, R); }
return res;
}
};
可持久化线段树
可持久化数组
template <int maxn, typename val_t>
struct PresistentArray {
struct Node {
int ls, rs;
val_t val;
} tr[maxn << 5];
int tot, root[maxn];
inline int newNode (Node old = Node { 0, 0, 0 }) { tr[++tot] = old; return tot; }
void build (int &now, int l, int r) {
now = newNode ();
if (l == r) { tr[now].val = a[l]; return; }
int mid = (l + r) >> 1;
build (tr[now].ls, l, mid);
build (tr[now].rs, mid + 1, r);
}
void modify (int &now, int old, int l, int r, int pos, val_t val) {
now = newNode (tr[old]);
if (l == r) { tr[now].val = val; return; }
int mid = (l + r) >> 1;
if (pos <= mid) { modify (tr[now].ls, tr[old].ls, l, mid, pos, val); }
if (pos > mid) { modify (tr[now].rs, tr[old].rs, mid + 1, r, pos, val); }
}
val_t query (int now, int l, int r, int pos) {
if (not now) { return 0; }
if (l == r) { return tr[now].val; }
int mid = (l + r) >> 1;
if (pos <= mid) { return query (tr[now].ls, l, mid, pos); }
if (pos > mid) { return query (tr[now].rs, mid + 1, r, pos); }
return 0;
}
};
01-trie
普通 01-trie
其实就是普通 01-trie
struct Trie {
i64 tr[maxn << 5][2];
int tot;
int siz[maxn << 5];
void insert (i64 val) {
int now = 0;
for (int i = 31; i >= 0; i--) {
bool x = ((val >> i) & 1);
if (not tr[now][x]) { tr[now][x] = ++tot; }
now = tr[now][x];
siz[now]++;
}
}
void del (i64 val) {
int now = 0;
for (int i = 31; i >= 0; i--) {
bool x = ((val >> i) & 1);
now = tr[now][x];
siz[now]--;
}
}
i64 query (i64 val) { // 使得结果尽可能小
int now = 0;
i64 ans = 0;
for (int i = 31; i >= 0; i--) {
bool x = ((val >> i) & 1);
if (not siz[tr[now][x]]) { ans |= (1ll << i); now = tr[now][x ^ 1]; }
else { now = tr[now][x]; }
}
return ans;
}
} trie;
查询 k 大 01-trie
与平衡树 / 值域线段树很像
struct Trie {
int tr[maxn << 5][2], tot, root;
int siz[maxn << 5];
Trie () { tot = 0; root = ++tot; }
inline void insert (i64 val) {
int now = root; siz[now] ++;
for (i64 i = 31; i >= 0; i--) {
bool x = ((val >> i) & 1ll);
if (not tr[now][x]) { tr[now][x] = ++tot; }
now = tr[now][x];
siz[now] ++;
}
}
inline i64 query (i64 val, int rnk) {
i64 res = 0;
int now = root;
for (i64 i = 31; i >= 0; i--) {
bool x = ((val >> i) & 1ll);
if (not tr[now][x ^ 1]) { now = tr[now][x]; continue; }
if (rnk <= siz[tr[now][x ^ 1]]) {
res |= 1ll << i;
now = tr[now][x ^ 1];
} else {
rnk -= siz[tr[now][x ^ 1]];
now = tr[now][x];
}
}
return res;
}
} trie;
可持久化 01-trie
struct Trie {
int tr[maxn << 6][2];
int tot;
int siz[maxn << 6];
int rt[maxn];
inline void clear () {
for (int i = 1; i <= tot; i++) { tr[i][0] = tr[i][1] = 0; siz[i] = 0; }
tot = 0;
}
inline void insert (int now, int old, int val) {
for (int i = 31; i >= 0; i--) {
bool x = ((val >> i) & 1);
tr[now][x] = ++tot; tr[now][x ^ 1] = tr[old][x ^ 1]; // clone
now = tr[now][x]; old = tr[old][x];
siz[now] = siz[old] + 1;
}
}
inline int query (int now, int old, int val) {
int ans = 0;
for (int i = 31; i >= 0; i--) {
bool x = ((val >> i) & 1);
if (siz[tr[now][x ^ 1]] - siz[tr[old][x ^ 1]] > 0) {
ans |= (1 << i);
now = tr[now][x ^ 1]; old = tr[old][x ^ 1];
} else {
now = tr[now][x]; old = tr[old][x];
}
}
return ans;
}
} trie;
平衡树
FHQ-Treap
普通平衡树
template <int maxn, typename val_t>
struct FhqTreap {
struct Node {
int ls, rs;
val_t val;
unsigned long long key;
int size;
} tr[maxn];
std::mt19937 rnd;
int tot, root;
inline int newNode (int val) {
tr[++tot] = Node { 0, 0, val, rnd (), 1 };
return tot;
}
inline void update (int now) { tr[now].size = tr[tr[now].ls].size + tr[tr[now].rs].size + 1; }
FhqTreap () { tot = 0; }
void split (int now, val_t val, int &x, int &y) {
if (not now) {
x = 0; y = 0;
} else if (tr[now].val <= val) {
x = now; split (tr[now].rs, val, tr[now].rs, y);
update (x);
} else {
y = now; split (tr[now].ls, val, x, tr[now].ls);
update (y);
}
}
int merge (int x, int y) {
if (not x or not y) {
return x | y;
} else if (tr[x].key < tr[y].key) {
tr[x].rs = merge (tr[x].rs, y);
update (x);
return x;
} else {
tr[y].ls = merge (x, tr[y].ls);
update (y);
return y;
}
}
inline void insert (val_t val) {
int x, y; split (root, val, x, y);
root = merge (merge (x, newNode (val)), y);
}
inline void remove (val_t val) {
int x, y, z;
split (root, val, x, y);
split (x, val - 1, x, z);
z = merge (tr[z].ls, tr[z].rs);
root = merge (merge (x, z), y);
}
inline val_t pre (val_t val) {
int x, y; split (root, val - 1, x, y);
int now = x;
while (tr[now].rs) { now = tr[now].rs; }
root = merge (x, y);
return tr[now].val;
}
inline val_t nxt (val_t val) {
int x, y; split (root, val, x, y);
int now = y;
while (tr[now].ls) { now = tr[now].ls; }
root = merge (x, y);
return tr[now].val;
}
inline int get_rank (val_t val) {
int x, y; split (root, val - 1, x, y);
int rank = tr[x].size + 1;
root = merge (x, y);
return rank;
}
inline val_t get_val (int rank) {
int now = root;
while (true) {
if (tr[tr[now].ls].size + 1 == rank) { return tr[now].val; }
else if (tr[tr[now].ls].size >= rank) { now = tr[now].ls; }
else { rank -= tr[tr[now].ls].size + 1; now = tr[now].rs; }
}
return -1;
}
};
文艺平衡树
其实就是维护中序遍历
struct Node {
int val;
int ls, rs;
int siz;
bool tag;
i64 key;
} tr[maxn];
int root, tot;
std::mt19937 rnd (std::chrono::steady_clock::now ().time_since_epoch ().count ());
inline int newNode (int val) {
tr[++tot] = Node { val, 0, 0, 1, false, rnd () };
return tot;
}
inline void pushUp (int now) {
tr[now].siz = tr[tr[now].ls].siz + tr[tr[now].rs].siz + 1;
}
inline void pushDown (int now) {
if (tr[now].tag) {
std::swap (tr[now].ls, tr[now].rs);
tr[tr[now].ls].tag ^= 1; tr[tr[now].rs].tag ^= 1;
tr[now].tag = false;
}
}
void split (int now, int siz, int& x, int& y) {
if (not now) { x = y = 0; return; }
pushDown (now);
if (siz <= tr[tr[now].ls].siz) {
y = now; split (tr[now].ls, siz, x, tr[now].ls);
pushUp (now);
} else {
x = now; split (tr[now].rs, siz - tr[tr[now].ls].siz - 1, tr[now].rs, y);
pushUp (now);
}
}
int merge (int x, int y) {
if (not x or not y) { return x | y; }
if (tr[x].key < tr[y].key) {
pushDown (x);
tr[x].rs = merge (tr[x].rs, y);
pushUp (x);
return x;
} else {
pushDown (y);
tr[y].ls = merge (x, tr[y].ls);
pushUp (y);
return y;
}
}
inline void rev (int l, int r) {
int x, y, z;
split (root, r, x, z);
split (x, l - 1, x, y);
tr[y].tag ^= 1;
root = merge (merge (x, y), z);
}
void output (int now) {
if (not now) { return; }
pushDown (now);
output (tr[now].ls);
std::cout << tr[now].val << " ";
output (tr[now].rs);
}
可持久化平衡树
其实跟可持久化线段树本质相同
似乎算导上讲过数据结构可持久化的通用方法
struct Node {
int val;
int ls, rs;
int siz;
unsigned long long key;
} tr[maxn * 50];
std::mt19937 rnd (std::chrono::steady_clock::now ().time_since_epoch ().count ());
int tot, root[maxn];
inline int newNode (Node old) {
tr[++tot] = old;
return tot;
}
inline void pushUp (int now) { tr[now].siz = tr[tr[now].ls].siz + tr[tr[now].rs].siz + 1; }
void split (int now, int val, int& x, int& y) {
if (not now) { x = y = 0; return; }
if (tr[now].val <= val) {
int id = newNode (tr[now]);
x = id; split (tr[id].rs, val, tr[id].rs, y);
pushUp (id);
} else {
int id = newNode (tr[now]);
y = id; split (tr[id].ls, val, x, tr[id].ls);
pushUp (id);
}
}
int merge (int x, int y) {
if (not x or not y) { return x | y; }
if (tr[x].key <= tr[y].key) {
int id = newNode (tr[x]);
tr[id].rs = merge (tr[id].rs, y);
pushUp (id);
return id;
} else {
int id = newNode (tr[y]);
tr[id].ls = merge (x, tr[id].ls);
pushUp (id);
return id;
}
}
inline void insert (int val, int i, int old) {
root[i] = root[old];
int x, y; split (root[i], val, x, y);
root[i] = merge (merge (x, newNode ({ val, 0, 0, 1, rnd () })), y);
}
inline void remove (int val, int i, int old) {
root[i] = root[old];
int x, y; split (root[i], val, x, y);
int z; split (x, val - 1, x, z);
z = merge (tr[z].ls, tr[z].rs);
root[i] = merge (merge (x, z), y);
}
inline int getRank (int val, int i, int old) {
root[i] = root[old];
int x, y; split (root[i], val - 1, x, y);
int rank = tr[x].siz + 1;
root[i] = merge (x, y);
return rank;
}
inline int getNum (int rank, int i, int old) {
root[i] = root[old];
int now = root[i];
while (now) {
if (tr[tr[now].ls].siz + 1 == rank) { return tr[now].val; }
if (rank <= tr[tr[now].ls].siz) { now = tr[now].ls; }
else { rank -= tr[tr[now].ls].siz + 1; now = tr[now].rs; }
}
}
inline int pre (int val, int i, int old) {
root[i] = root[old];
int x, y; split (root[i], val - 1, x, y);
int now = x;
while (tr[now].rs) { now = tr[now].rs; }
root[i] = merge (x, y);
return now ? tr[now].val : -2147483647;
}
inline int nxt (int val, int i, int old) {
root[i] = root[old];
int x, y; split (root[i], val, x, y);
int now = y;
while (tr[now].ls) { now = tr[now].ls; }
root[i] = merge (x, y);
return now ? tr[now].val : 2147483647;
}
树套树
树状数组套线段树
struct FenwickTree {
struct SegmentTree {
struct Node {
int ls, rs;
int sum;
} tr[maxn << 9];
std::vector<int> Add;
std::vector<int> Sub;
int root[maxn], tot;
void modify (int& now, int l, int r, int pos, int val) {
if (not now) { now = ++tot; }
tr[now].sum += val;
if (l == r) { return; }
int mid = (l + r) >> 1;
if (pos <= mid) { modify (tr[now].ls, l, mid, pos, val); }
if (pos > mid) { modify (tr[now].rs, mid + 1, r, pos, val); }
}
int query (int l, int r, int k) {
if (l == r) { return l; }
int sum = 0;
for (auto x : Add) { sum += tr[tr[x].ls].sum; }
for (auto x : Sub) { sum -= tr[tr[x].ls].sum; }
for (auto& x : Add) { x = (k <= sum) ? tr[x].ls : tr[x].rs; }
for (auto& x : Sub) { x = (k <= sum) ? tr[x].ls : tr[x].rs; }
int mid = (l + r) >> 1;
return (k <= sum) ? query (l, mid, k) : query (mid + 1, r, k - sum);
}
} seg;
inline int lowbit (int x) { return x & (-x); }
inline void ins (int pos, int val) {
for (int i = pos; i <= n; i += lowbit (i)) { seg.modify (seg.root[i], 0, 1e9, val, 1); }
}
inline void del (int pos, int val) {
for (int i = pos; i <= n; i += lowbit (i)) { seg.modify (seg.root[i], 0, 1e9, val, -1); }
}
inline int query (int l, int r, int k) {
seg.Add.clear (); for (int i = r; i; i -= lowbit (i)) { seg.Add.push_back (seg.root[i]); }
seg.Sub.clear (); for (int i = l - 1; i; i -= lowbit (i)) { seg.Sub.push_back (seg.root[i]); }
return seg.query (0, 1e9, k);
}
} tr;
笛卡尔树
神秘科技
inline int build () {
std::stack <int> rChain; // 右链
rep (now, 1, n) {
int last = 0;
while (not rChain.empty () and w[rChain.top ()] > w[now]) { last = rChain.top (); rChain.pop (); }
tr[rChain.empty () ? 0 : rChain.top ()].rs = now;
tr[now].ls = last;
rChain.push (now);
}
return tr[0].rs;
}
树
树链剖分
核心部分代码不长
int fa[maxn], siz[maxn], son[maxn], dep[maxn];
void build_tree (int now) {
siz[now] = 1;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (t == fa[now]) { continue; }
fa[t] = now;
dep[t] = dep[now] + 1;
build_tree (t);
siz[now] += siz[t];
if (siz[t] > siz[son[now]]) { son[now] = t; }
}
}
int top[maxn], dfn[maxn], dpos, rnk[maxn];
// int bottom[maxn];
void tree_decomp (int now, int topnow) {
dfn[now] = ++dpos;
rnk[dfn[now]] = now;
top[now] = topnow;
// bottom[now] = dfn[now];
if (not son[now]) { return; }
tree_decomp (son[now], topnow);
// bottom[now] = std::max (bottom[now], bottom[son[now]]);
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (t == fa[now]) { continue; }
if (t == son[now]) { continue; }
tree_decomp (t, t);
// bottom[now] = std::max (bottom[now], bottom[t]);
}
}
子树内的信息
一般会像上面注释里面一样维护一个 bottom[u] 表示以 \(u\) 为根的子树中最大的 \(dfn\)
然后就可以
inline i64 subtreeQuery (int x) {
return tr.query (1, 1, n, dfn[x], bottom[x]);
}
修改同理
路径上的信息
直接跳链即可
inline void chainModify (int x, int y, i64 val) {
while (top[x] != top[y]) {
if (dep[top[x]] > dep[top[y]]) {
tr.modify (1, 1, n, dfn[top[x]], dfn[x], val);
x = fa[top[x]];
} else {
tr.modify (1, 1, n, dfn[top[y]], dfn[y], val);
y = fa[top[y]];
}
}
if (dfn[x] > dfn[y]) { std::swap (x, y); }
tr.modify (1, 1, n, dfn[x], dfn[y], val);
}
查询同理
lca
同样是直接跳链
inline int lca (int u, int v) {
while (top[u] != top[v]) {
if (dep[top[u]] > dep[top[v]]) {
u = fa[top[u]];
} else {
v = fa[top[v]];
}
}
return dep[u] < dep[v] ? u : v;
}
Link-Cut Tree
struct Node {
int sum, val;
int ch[2];
int fa;
bool tag;
} tr[maxn];
inline bool chk (int now) { return now == tr[tr[now].fa].ch[1]; }
inline bool notRoot (int now) { return now == tr[tr[now].fa].ch[0] or now == tr[tr[now].fa].ch[1]; }
inline void pushUp (int now) { tr[now].sum = tr[tr[now].ch[0]].sum xor tr[tr[now].ch[1]].sum xor tr[now].val; }
inline void rev (int now) { std::swap (tr[now].ch[0], tr[now].ch[1]); tr[now].tag ^= 1; }
inline void pushDown (int now) {
if (not tr[now].tag) { return; }
if (tr[now].ch[0]) { rev (tr[now].ch[0]); }
if (tr[now].ch[1]) { rev (tr[now].ch[1]); }
tr[now].tag = false;
}
void pushAll (int now) {
if (notRoot (now)) { pushAll (tr[now].fa); }
pushDown (now);
}
inline void connect (int fa, int x, int p) { tr[fa].ch[p] = x; tr[x].fa = fa; }
inline void rotate (int x) {
int f = tr[x].fa, ff = tr[f].fa, p = chk (x), pp = chk (f);
pushDown (f); pushDown (x);
if (notRoot (f)) { connect (ff, x, pp); }
else { tr[x].fa = ff; }
connect (f, tr[x].ch[p ^ 1], p); connect (x, f, p ^ 1);
pushUp (f); pushUp (x);
}
inline void splay (int x) {
pushAll (x);
for (; notRoot (x); rotate (x)) {
if (notRoot (tr[x].fa)) { rotate (chk (tr[x].fa) == chk (x) ? tr[x].fa : x); }
}
pushUp (x);
}
inline void access (int x) {
int pre = 0;
while (x) {
splay (x); tr[x].ch[1] = pre; pushUp (x);
x = tr[pre = x].fa;
}
}
inline void makeRoot (int x) { access (x); splay (x); rev (x); }
inline void split (int x, int y) { makeRoot (x); access (y); splay (y); }
inline int findRoot (int x) {
access (x); splay (x);
pushDown (x); while (tr[x].ch[0]) { x = tr[x].ch[0]; pushDown (x); }
splay (x); return x;
}
inline void link (int x, int y) {
makeRoot (x);
if (x == findRoot (y)) { return; }
tr[x].fa = y;
}
inline void cut (int x, int y) {
split (x, y);
if (tr[y].ch[0] != x or tr[x].ch[0] != tr[x].ch[1]) { return; }
tr[y].ch[0] = tr[x].fa = 0;
pushUp (y);
}
点分治
淀粉质
inline void addEdge (int u, int v, int w) {
es[++cnt] = Edge { u, v, last[u], w };
last[u] = cnt;
}
int sum;
std::bitset<maxn> removed;
int root;
int maxpart[maxn];
int siz[maxn];
void getRoot (int now, int fa) {
siz[now] = 1;
maxpart[now] = 0;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (t == fa) { continue; }
if (removed[t]) { continue; }
getRoot (t, now);
siz[now] += siz[t];
maxpart[now] = std::max (maxpart[now], siz[t]);
}
maxpart[now] = std::max (maxpart[now], sum - siz[now]);
if (maxpart[now] < maxpart[root]) { root = now; }
}
int disStack[maxn], dpos;
int removeStack[maxn], rpos;
int dis[maxn];
void getDis (int now, int fa) {
disStack[++dpos] = dis[now];
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (t == fa) { continue; }
if (removed[t]) { continue; }
dis[t] = dis[now] + es[i].w;
getDis (t, now);
}
}
const int maxv = 100000005;
std::bitset<maxv> exist;
void solve (int now) {
exist[0] = true;
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (removed[t]) { continue; }
dis[t] = es[i].w;
getDis (t, now);
for (int j = 1; j <= dpos; j++) {
for (auto &q : qs) {
if (q.k - disStack[j] >= 0) { q.ans |= exist[q.k - disStack[j]]; }
}
}
while (dpos) {
int top = disStack[dpos--];
exist[top] = true;
removeStack[++rpos] = top;
}
}
while (rpos) {
int top = removeStack[rpos--];
exist[top] = false;
}
}
inline void divide (int now) {
removed[now] = true;
solve (now);
for (int i = last[now]; i; i = es[i].pre) {
int t = es[i].v;
if (removed[t]) { continue; }
sum = siz[t];
maxpart[root] = 0x3f3f3f3f;
getRoot (t, now);
getRoot (root, now); // 实际上在getSiz
divide (root);
}
}
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