20250819
T1
集合查询
\(x | \gcd(v, k) \longleftrightarrow x | v \land x | k\)，对每个数维护一棵 01trie，加入一个数时加入其所有因数的 01trie 即可。要求和不超过的限制是容易的，每次要进入 trie 的一棵子树时判断子树 \(\max\) 是否超过限制即可。
代码
#include <iostream>
#include <string.h>
#include <vector>
using namespace std;
int q;
vector<int> vec[100005];
int rt[100005], ncnt;
int son[20000005][2];
int mn[20000005];
void Insert(int x, int y) {
int p = rt[x];
if (!p) p = rt[x] = ++ncnt;
mn[p] = min(mn[p], y);
for (int i = 18; ~i; i--) {
int t = ((y >> i) & 1);
if (!son[p][t]) son[p][t] = ++ncnt;
p = son[p][t];
mn[p] = min(mn[p], y);
}
}
int Query(int p, int x, int s) {
int y = 0;
for (int i = 18; ~i; i--) {
int t = (!((x >> i) & 1));
if (son[p][t] && x + mn[son[p][t]] <= s) y |= (t << i), p = son[p][t];
else y |= ((!t) << i), p = son[p][!t];
}
return y;
}
bool vis[100005];
int main() {
freopen("set.in", "r", stdin);
freopen("set.out", "w", stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
memset(mn, 63, sizeof mn);
for (int i = 1; i <= 100000; i++) {
for (int j = i; j <= 100000; j += i)
vec[j].emplace_back(i);
}
cin >> q;
while (q--) {
int op, x, y, z;
cin >> op >> x;
if (op == 1) {
if (vis[x]) continue;
vis[x] = 1;
for (int v : vec[x]) Insert(v, x);
} else {
cin >> y >> z;
if (mn[rt[y]] + x > z || x % y != 0) { cout << "-1\n"; continue; }
cout << Query(rt[y], x, z) << "\n";
}
}
return 0;
}
T2
公交路线
每次跳到能跳到的最高的位置，直到将要跳过 LCA。然后查询是否存在一条线路两端分别位于当前两个点的子树中即可。前面的可以树上倍增，后面的直接二维数点。可能需要判掉询问点有祖先后代关系的情况。
代码
#include <iostream>
#include <algorithm>
#include <string.h>
#include <array>
#define lowbit(x) ((x) & (-(x)))
using namespace std;
const int inf = 0x3f3f3f3f;
inline void Cmin(int &x, int y) { x = min(x, y); }
int n, m, K;
int head[100005], nxt[200005], to[200005], ecnt;
void add(int u, int v) { to[++ecnt] = v, nxt[ecnt] = head[u], head[u] = ecnt; }
int top[100005], son[100005], dep[100005], sz[100005], f[100005];
int L[100005], R[100005], ncnt;
int _dfn[100005];
int F[100005][20];
void dfs1(int x, int fa, int d) {
dep[x] = d;
f[x] = fa;
sz[x] = 1;
for (int i = head[x]; i; i = nxt[i]) {
int v = to[i];
if (v != fa) {
dfs1(v, x, d + 1);
sz[x] += sz[v];
if (sz[v] > sz[son[x]])
son[x] = v;
}
}
}
void dfs2(int x, int t) {
top[x] = t;
_dfn[L[x] = ++ncnt] = x;
if (son[x]) dfs2(son[x], t);
for (int i = head[x]; i; i = nxt[i]) {
int v = to[i];
if (v != f[x] && v != son[x])
dfs2(v, v);
}
R[x] = ncnt;
}
int LCA(int x, int y) {
while (top[x] ^ top[y]) (dep[top[x]] < dep[top[y]]) ? (y = f[top[y]]) : (x = f[top[x]]);
return (dep[x] < dep[y] ? x : y);
}
int up[100005], stk[100005];
array<int, 3> ln[100005];
void dfs3(int x) {
stk[dep[x]] = x;
for (int i = head[x]; i; i = nxt[i]) {
int v = to[i];
if (v != f[x]) {
dfs3(v);
Cmin(up[x], up[v]);
}
}
F[x][0] = stk[up[x]];
}
void dfs4(int x) {
for (int i = 1; i < 20; i++) F[x][i] = F[F[x][i - 1]][i - 1];
for (int i = head[x]; i; i = nxt[i]) { int v = to[i]; if (v != f[x]) dfs4(v); }
}
int s[100005], t[100005];
struct Query {
int x, y, id, v;
} qs[500005];
int qcnt;
int tmp[100005];
struct BIT {
int bit[100005];
void add(int x, int y) { for (; x <= 100000; x += lowbit(x)) bit[x] += y; }
int query(int x) {
int ret = 0;
for (; x; x -= lowbit(x)) ret += bit[x];
return ret;
}
} bit;
int ans[100005];
int goUp(int &x, int y) {
if (dep[F[x][19]] > dep[y]) return inf;
int cnt = 0;
for (int i = 19; ~i; i--) if (dep[F[x][i]] > dep[y]) x = F[x][i], cnt += (1 << i);
return cnt + 1;
}
int main() {
freopen("bus.in", "r", stdin);
freopen("bus.out", "w", stdout);
cin >> n >> m >> K;
for (int i = 1; i < n; i++) {
int u, v;
cin >> u >> v;
add(u, v);
add(v, u);
}
dfs1(1, 0, 1);
dfs2(1, 1);
for (int i = 1; i <= n; i++) up[i] = dep[i];
for (int i = 1; i <= m; i++) {
cin >> ln[i][0] >> ln[i][1];
if (L[ln[i][0]] > L[ln[i][1]]) swap(ln[i][0], ln[i][1]);
ln[i][2] = LCA(ln[i][0], ln[i][1]);
Cmin(up[ln[i][0]], dep[ln[i][2]]);
Cmin(up[ln[i][1]], dep[ln[i][2]]);
qs[++qcnt] = (Query) { L[ln[i][0]], L[ln[i][1]], -1, 1 };
}
dfs3(1);
dfs4(1);
for (int i = 1; i <= K; i++) {
cin >> s[i] >> t[i];
(dep[s[i]] < dep[t[i]]) ? void() : swap(s[i], t[i]);
if (L[s[i]] <= L[t[i]] && L[t[i]] <= R[s[i]]) ans[i] = goUp(t[i], s[i]);
else {
int z = LCA(s[i], t[i]);
ans[i] = goUp(s[i], z) + goUp(t[i], z);
if (L[s[i]] > L[t[i]]) swap(s[i], t[i]);
qs[++qcnt] = (Query) { L[s[i]] - 1, L[t[i]] - 1, i, 1 };
qs[++qcnt] = (Query) { R[s[i]], L[t[i]] - 1, i, -1 };
qs[++qcnt] = (Query) { L[s[i]] - 1, R[t[i]], i, -1 };
qs[++qcnt] = (Query) { R[s[i]], R[t[i]], i, 1 };
}
}
sort(qs + 1, qs + qcnt + 1, [](Query x, Query y) { return x.x == y.x ? (x.y == y.y ? (x.id < y.id) : (x.y < y.y)) : (x.x < y.x); });
for (int i = 1; i <= qcnt; i++) {
if (qs[i].id < 0) bit.add(qs[i].y, 1);
else tmp[qs[i].id] += qs[i].v * bit.query(qs[i].y);
}
for (int i = 1; i <= K; i++) {
if (ans[i] >= inf - 1) cout << "-1\n";
else cout << ans[i] - (tmp[i] != 0) - 1 << "\n";
}
return 0;
}
T3
游戏
先求出每条边上剩下的堆异或和为每个数的概率，然后跑一遍矩阵树即可。可以先 fwt 然后扔进矩阵树最后再 ifwt，这样中途就不用来回变了。接下来要求出剩下的异或和为每个数的概率，实际上是要把一车形如 \(\frac{3}{4}x^{\varnothing} + \frac{1}{4}x^{S}\) 的东西卷起来。观察卷完之后的 fwt 数组，发现只会有 \(1\) 和 \(\frac{1}{2}\) 两种值，由异或 fwt 的定义容易发现，且 \(\frac{1}{2}\) 只会在 \(x^{T}(|T \cap S| \equiv 1 \bmod 2)\) 的情况下出现。因此只需要求有多少数和当前位置指数上的集合交大小为奇数。我们考虑对计数数组进行 fwt，这样每个位置上的数就是偶数 - 奇数，而偶数 + 奇数就是所有数的总数。把两个减了除以二就可以算出奇数的个数。于是就做完了。
代码
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
#define int long long
using namespace std;
const int P = 998244353, i2 = (P + 1) / 2, i4 = i2 * i2 % P;
inline void Madd(int &x, int y) { (x += y) >= P ? (x -= P) : 0; }
inline int Msum(int x, int y) { return Madd(x, y), x; }
inline int qpow(int x, int y = P - 2) {
int ret = 1;
while (y) {
if (y & 1) ret = ret * x % P;
y >>= 1; x = x * x % P;
}
return ret;
}
int n, m;
struct poly {
vector<int> vec;
int& operator[](int x) { return vec[x]; }
poly() { vec.resize(1 << 10); }
void operator+=(poly &b) { for (int i = 0; i < (1 << m); i++) Madd(vec[i], b[i]); }
void operator-=(poly b) { for (int i = 0; i < (1 << m); i++) Madd(vec[i], P - b[i]); }
void operator*=(poly &b) { for (int i = 0; i < (1 << m); i++) vec[i] = vec[i] * b[i] % P; }
poly operator~() { poly ret; for (int i = 0; i < (1 << m); i++) ret[i] = qpow(vec[i]); return ret; }
bool _0() { return *max_element(vec.begin(), vec.begin() + (1 << m)) == 0; }
} A[75][75];
inline poly operator*(poly a, poly &b) { return a *= b, a; }
void FWT(poly &f, int t = 1) {
for (int i = 1; i < (1 << m); i <<= 1) {
for (int j = 0; j < (1 << m); j += (i << 1)) {
for (int k = 0; k < i; k++) {
int x = f[j | k], y = f[i | j | k];
if (t == 1) {
f[j | k] = Msum(x, y);
f[i | j | k] = Msum(x, P - y);
} else {
f[j | k] = Msum(x, y) * i2 % P;
f[i | j | k] = Msum(x, P - y) * i2 % P;
}
}
}
}
}
poly detp(int N) {
int rc = 1;
for (int i = 1; i < N; i++) {
if (A[i][i]._0()) {
for (int j = i; j < N; j++) {
if (!A[j][i]._0()) {
rc = P - rc;
swap(A[i], A[j]);
break;
}
}
}
poly coe = ~A[i][i];
for (int j = i + 1; j < N; j++) {
if (!A[j][i]._0()) {
for (int k = N - 1; k >= i; k--)
A[j][k] -= A[i][k] * coe * A[j][i];
}
}
}
poly ret;
for (int i = 0; i < (1 << m); i++) ret[i] = rc;
for (int i = 1; i < N; i++) ret *= A[i][i];
return ret;
}
int iA[75][75];
int deti(int N) {
int ret = 1;
for (int i = 1; i < N; i++) {
if (iA[i][i] == 0) {
for (int j = i; j < N; j++) {
if (iA[j][i]) {
ret = P - ret;
swap(iA[i], iA[j]);
break;
}
}
}
int coe = qpow(iA[i][i]);
for (int j = 1; j < N; j++) {
if (i != j && iA[j][i]) {
for (int k = N; k >= i; k--)
Madd(iA[j][k], P - iA[i][k] * coe % P * iA[j][i] % P);
}
}
}
for (int i = 1; i < N; i++) ret = ret * iA[i][i] % P;
return ret;
}
signed main() {
freopen("gemo.in", "r", stdin);
freopen("gemo.out", "w", stdout);
cin >> n >> m;
for(int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
int K, tk, x; cin >> K; tk = K;
if (K) {
poly tmp0;
while (K--) cin >> x, ++tmp0[x];
FWT(tmp0);
for (int k = 0; k < (1 << m); k++) tmp0[k] = qpow(i2, (tk + P - tmp0[k]) * i2 % P);
Madd(iA[i][j], P - 1), Madd(iA[j][i], P - 1);
++iA[i][i], ++iA[j][j];
A[i][j] -= tmp0, A[j][i] -= tmp0;
A[i][i] += tmp0, A[j][j] += tmp0;
}
}
}
poly res = detp(n);
int ret = deti(n);
FWT(res, -1);
cout << (P + 1 - res[0] * qpow(ret) % P) % P << "\n";
return 0;
}
T4
音符大师
考虑一个朴素的 dp：\(f_{i, j, k}\) 表示当前打完了第 \(i\) 个，一个手在 \(j\)，另一个在 \(k\) 的最小代价。容易发现 \(j,k\) 至少一个 \(=i\)，于是把那两维合并，剩下一维记不在 \(i\) 的那只手在哪。显然如果下一个音符落在当前手的覆盖范围内，那肯定不会动的。于是第一维只需要记 \(0 / 1\) 表示当前在 \(p_i\) 还是 \(p_i - L\)。另一维也是同理，只需要记 \(2n\) 个位置。线段树维护，每次转移分类讨论：下一个音符拿当前在 \(i\) 的手打，那么找到下一个不在 \([p_i, p_i + L]\) 的音符 \(x\)（找不到直接更新答案），把当前线段树打个全局加 tag 合并过去即可。若下一个音符拿另一个手打，那么还是考虑 \(x\) 能否被另一个手打到，如果不能，就把当前的 \(f_{i, j}\) 更新到 \(f_{x, i}\) 上去（注意 \(i\) 从第一维 变成了第二维，因为它变成“另一只手”了）；如果能打到，那么这样的 \(j\) 只有至多 \(L\) 个，再找到第一个不被两个手包含的音符，考虑这个音符要拿哪个手打，分别做单点更新即可。总时间复杂度 \(\mathcal{O}(nL\log n)\)。不知道出于什么心理要卡正解常。
代码
#pragma GCC optimize("Ofast,inline,unroll-loops")
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const int _inf = 0x3f3f3f3f;
inline void Cmin(long long &x, long long y) { x = min(x, y); }
#define getchar() p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++
char buf[1<<21], *p1, *p2, ch;
long long read() {
long long ret = 0, neg = 0; char c = getchar(); neg = (c == '-');
while (c < '0' || c > '9') c = getchar(), neg |= (c == '-');
while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
return ret * (neg ? -1 : 1);
}
int n, m;
long long ans = inf;
int a[50005], d1[50005], dcnt1;
int d2[100005], dcnt2;
int rt[50005];
struct Persistent_Segment_Tree {
struct node { int l, r, mn; } T[1500005];
int ncnt;
void Insert(int &p, int q, int l, int r, int x, int y) {
T[p = ++ncnt] = T[q];
T[p].mn = y;
if (l == r) return;
int mid = (l + r) >> 1;
if (x <= mid) Insert(T[p].l, T[q].l, l, mid, x, y);
else Insert(T[p].r, T[q].r, mid + 1, r, x, y);
}
int Query(int o, int l, int r, int L, int R) {
if (L > R || !o) return _inf;
if (L <= l && r <= R) return T[o].mn;
int mid = (l + r) >> 1;
if (R <= mid) return Query(T[o].l, l, mid, L, R);
if (L > mid) return Query(T[o].r, mid + 1, r, L, R);
return min(Query(T[o].l, l, mid, L, R), Query(T[o].r, mid + 1, r, L, R));
}
} pseg;
struct Segment_Tree {
struct node {
int l, r;
long long tg, mn1, mn2;
} T[7000005];
int ncnt;
inline void tag(int o, long long v) { T[o].tg += v, T[o].mn1 += v, T[o].mn2 += v; }
inline void pushup(int o) { T[o].mn1 = min(T[T[o].l].mn1, T[T[o].r].mn1), T[o].mn2 = min(T[T[o].l].mn2, T[T[o].r].mn2); }
void pushdown(int o) {
if (!T[o].tg) return;
T[o].l ? tag(T[o].l, T[o].tg) : void();
T[o].r ? tag(T[o].r, T[o].tg) : void();
T[o].tg = 0;
}
void Insert(int &o, int l, int r, int x, long long y) {
if (!o) T[o = ++ncnt] = (node) { 0, 0, 0, inf, inf };
if (l == r) {
T[o].mn1 = min(T[o].mn1, y - d2[x]);
T[o].mn2 = min(T[o].mn2, y + d2[x]);
return;
}
pushdown(o);
int mid = (l + r) >> 1;
if (x <= mid) Insert(T[o].l, l, mid, x, y);
else Insert(T[o].r, mid + 1, r, x, y);
pushup(o);
}
int Merge(int p, int q, int l, int r) {
if (!p || !q) return p | q;
if (l == r) {
T[p].mn1 = min(T[p].mn1, T[q].mn1), T[p].mn2 = min(T[p].mn2, T[q].mn2);
return p;
}
pushdown(p), pushdown(q);
int mid = (l + r) >> 1;
T[p].l = Merge(T[p].l, T[q].l, l, mid);
T[p].r = Merge(T[p].r, T[q].r, mid + 1, r);
pushup(p);
return p;
}
long long Query1(int o, int l, int r, int L, int R) {
if (!o) return inf;
if (L <= l && r <= R) return T[o].mn1;
int mid = (l + r) >> 1;
pushdown(o);
if (R <= mid) return Query1(T[o].l, l, mid, L, R);
if (L > mid) return Query1(T[o].r, mid + 1, r, L, R);
return min(Query1(T[o].l, l, mid, L, R), Query1(T[o].r, mid + 1, r, L, R));
}
long long Query2(int o, int l, int r, int L, int R) {
if (!o) return inf;
if (L <= l && r <= R) return T[o].mn2;
int mid = (l + r) >> 1;
pushdown(o);
if (R <= mid) return Query2(T[o].l, l, mid, L, R);
if (L > mid) return Query2(T[o].r, mid + 1, r, L, R);
return min(Query2(T[o].l, l, mid, L, R), Query2(T[o].r, mid + 1, r, L, R));
}
void get(int o, int l, int r, int L, int R, vector<pair<int, long long> > &vec) {
if (!o) return;
if (l == r) return vec.emplace_back(l, T[o].mn2), void();
int mid = (l + r) >> 1;
pushdown(o);
if (R <= mid) return get(T[o].l, l, mid, L, R, vec);
if (L > mid) return get(T[o].r, mid + 1, r, L, R, vec);
return get(T[o].l, l, mid, L, R, vec), get(T[o].r, mid + 1, r, L, R, vec);
}
} seg;
int rt2[50005][2];
long long mn[50005][2];
signed main() {
seg.T[0] = (Segment_Tree::node) { 0, 0, 0, inf, inf };
freopen("music.in", "r", stdin);
freopen("music.out", "w", stdout);
memset(mn, 63, sizeof mn);
n = read(), m = read();
for (int i = 1; i <= n; i++) a[i] = read(), d1[i] = a[i], d2[++dcnt2] = a[i], d2[++dcnt2] = a[i] - m;
d1[dcnt1 = n + 1] = 0; d2[++dcnt2] = 0; d2[++dcnt2] = -m;
sort(d1 + 1, d1 + dcnt1 + 1);
sort(d2 + 1, d2 + dcnt2 + 1);
dcnt1 = unique(d1 + 1, d1 + dcnt1 + 1) - d1 - 1;
dcnt2 = unique(d2 + 1, d2 + dcnt2 + 1) - d2 - 1;
for (int i = n; i; i--) {
int x = lower_bound(d1 + 1, d1 + dcnt1 + 1, a[i]) - d1;
pseg.Insert(rt[i], rt[i + 1], 1, dcnt1, x, i);
}
#define _l1(x) lower_bound(d1 + 1, d1 + dcnt1 + 1, x) - d1
#define _u1(x) upper_bound(d1 + 1, d1 + dcnt1 + 1, x) - d1
#define _l2(x) lower_bound(d2 + 1, d2 + dcnt2 + 1, x) - d2
#define _u2(x) upper_bound(d2 + 1, d2 + dcnt2 + 1, x) - d2
seg.Insert(rt2[0][0], 1, dcnt2, _l2(0), 0);
mn[0][0] = 0;
for (int i = 0; i <= n; i++) for (int _ : { 0, 1 }) {
int tl = a[i] - _ * m, tr = a[i] + (!_) * m, x = _l2(tl), t1 = _l1(tl), t2 = _u1(tr), y = min(pseg.Query(rt[i + 1], 1, dcnt1, 1, t1 - 1), pseg.Query(rt[i + 1], 1, dcnt1, t2, dcnt1));
int v1 = pseg.Query(rt[i + 1], 1, dcnt1, 1, t1 - 1), v2 = pseg.Query(rt[i + 1], 1, dcnt1, t2, dcnt1);
long long tmp;
if (y >= _inf) { Cmin(ans, mn[i][_]); continue; }
tmp = seg.Query2(rt2[i][_], 1, dcnt2, _u2(a[y]), dcnt2) - a[y];
seg.Insert(rt2[y][0], 1, dcnt2, x, tmp); Cmin(mn[y][0], tmp);
tmp = a[y] - m + seg.Query1(rt2[i][_], 1, dcnt2, 1, _l2(a[y] - m) - 1);
seg.Insert(rt2[y][1], 1, dcnt2, x, tmp); Cmin(mn[y][1], tmp);
if (a[y] < tl) {
vector<pair<int, long long> > vec;
seg.get(rt2[i][_], 1, dcnt2, _l2(a[y] - m), _u2(a[y]) - 1, vec);
for (auto p : vec) {
int t = p.first, u = d2[t];
long long val = p.second;
if (val >= inf) continue;
int z = min({ pseg.Query(rt[i + 1], 1, dcnt1, 1, _l1(u) - 1), pseg.Query(rt[i + 1], 1, dcnt1, _u1(u + m), t1 - 1), v2 });
if (z >= _inf) { Cmin(ans, val - u); continue; }
if (a[z] < tl) seg.Insert(rt2[z][0], 1, dcnt2, t, tmp = val - u + tl - a[z]), Cmin(mn[z][0], tmp);
else seg.Insert(rt2[z][1], 1, dcnt2, t, tmp = val - u + a[z] - m - tl), Cmin(mn[z][1], tmp);
if (a[z] < u) seg.Insert(rt2[z][0], 1, dcnt2, x, tmp = val - a[z]), Cmin(mn[z][0], tmp);
else seg.Insert(rt2[z][1], 1, dcnt2, x, tmp = val - u - u + a[z] - m), Cmin(mn[z][1], tmp);
}
seg.tag(rt2[i][_], tl - a[y]);
rt2[y][0] = seg.Merge(rt2[y][0], rt2[i][_], 1, dcnt2);
Cmin(mn[y][0], mn[i][_] + tl - a[y]);
} else {
vector<pair<int, long long> > vec;
seg.get(rt2[i][_], 1, dcnt2, _l2(a[y] - m), _u2(a[y]) - 1, vec);
for (auto p : vec) {
int t = p.first, u = d2[t];
long long val = p.second;
if (val >= inf) continue;
int z = min({ v1, pseg.Query(rt[i + 1], 1, dcnt1, t2, _l1(u) - 1), pseg.Query(rt[i + 1], 1, dcnt1, _u1(u + m), dcnt1) });
if (z >= _inf) { Cmin(ans, val - u); continue; }
if (a[z] < tl) seg.Insert(rt2[z][0], 1, dcnt2, t, tmp = val - u + tl - a[z]), Cmin(mn[z][0], tmp);
else seg.Insert(rt2[z][1], 1, dcnt2, t, tmp = val - u + a[z] - m - tl), Cmin(mn[z][1], tmp);
if (a[z] < u) seg.Insert(rt2[z][0], 1, dcnt2, x, tmp = val - a[z]), Cmin(mn[z][0], tmp);
else seg.Insert(rt2[z][1], 1, dcnt2, x, tmp = val - u - u + a[z] - m), Cmin(mn[z][1], tmp);
}
seg.tag(rt2[i][_], a[y] - tl - m);
rt2[y][1] = seg.Merge(rt2[y][1], rt2[i][_], 1, dcnt2);
Cmin(mn[y][1], mn[i][_] + a[y] - tl - m);
}
}
cout << ans << "\n";
return 0;
}
T3 经典套路。T4 经典另一维线段树维护 dp。来回更新。
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