20250912
T1
序列
显然每次操作保证 \(a\) 单增是假的，一定可以构造合理的操作顺序无视掉这个限制。每个位置确定划分多少次之后一定均分最优，拿个堆维护把每个东西多分一段的收益，每次选择收益最大的即可。
代码
#include <iostream>
#include <string.h>
#include <queue>
#define int long long
using namespace std;
const int P = 998244353;
int n, m;
int a[100005], b[100005], c[100005];
int cur[100005];
int ans;
priority_queue<pair<int, int> > q;
int calc(int x, int y) {
x = c[x];
int r = x % y;
return (x / y) * (x / y) * (y - r) + (x / y + 1) * (x / y + 1) * r;
}
signed main() {
freopen("seq.in", "r", stdin);
freopen("seq.out", "w", stdout);
cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = 1; i <= n; i++) cin >> b[i];
for (int i = 1; i <= n; i++) c[i] = abs(a[i] - b[i]), ans += calc(i, cur[i] = 1) % P, q.push({ calc(i, 1) - calc(i, 2), i }), m -= (a[i] != b[i]);
ans %= P;
if (m < 0) return cout << "-1\n", 0;
while (m) {
pair<int, int> p = q.top();
q.pop();
ans += P - p.first % P;
--m;
int x = p.second;
++cur[x], q.push({ calc(x, cur[x]) - calc(x, cur[x] + 1), x });
}
cout << ans % P << "\n";
return 0;
}
T2
石头剪刀布
线段树板子。
代码
#include <iostream>
#include <string.h>
#include <array>
#include <map>
using namespace std;
int n, q;
string str;
array<int, 3> p[6] = {
{ 0, 1, 2 },
{ 0, 2, 1 },
{ 1, 0, 2 },
{ 1, 2, 0 },
{ 2, 0, 1 },
{ 2, 1, 0 }
};
inline int _p(array<int, 3> x) { return x[0] * 2 + (x[1] > x[2]); }
// 012, rsp
array<int, 3> operator+(array<int, 3> x, array<int, 3> y) { return { y[x[0]], y[x[1]], y[x[2]] }; }
array<int, 3> tmp[6];
int f(int x, int y) { return ((x + 1) % 3 == y) ? x : y; }
struct Segment_Tree {
array<int, 3> S[800005][6], tg[800005];
void tag(int o, array<int, 3> v) {
for (int i = 0; i < 6; i++) tmp[i] = S[o][i];
for (int i = 0; i < 6; i++) S[o][i] = tmp[_p(v + p[i])];
tg[o] = tg[o] + v;
}
void pushdown(int o) {
if (tg[o] == p[0]) return;
tag(o << 1, tg[o]);
tag(o << 1 | 1, tg[o]);
tg[o] = p[0];
}
void pushup(int o) { for (int i = 0; i < 6; i++) S[o][i] = S[o << 1][i] + S[o << 1 | 1][i]; }
void Build(int o, int l, int r) {
tg[o] = p[0];
if (l == r) {
int x = str[l] - 'A';
for (int i = 0; i < 6; i++) S[o][i] = { f(0, p[i][x]), f(1, p[i][x]), f(2, p[i][x]) };
return;
}
int mid = (l + r) >> 1;
Build(o << 1, l, mid);
Build(o << 1 | 1, mid + 1, r);
pushup(o);
}
void Change(int o, int l, int r, int L, int R, array<int, 3> v) {
if (L <= l && r <= R) return tag(o, v);
pushdown(o);
int mid = (l + r) >> 1;
if (L <= mid) Change(o << 1, l, mid, L, R, v);
if (R > mid) Change(o << 1 | 1, mid + 1, r, L, R, v);
pushup(o);
}
array<int, 3> Query(int o, int l, int r, int L, int R) {
if (L <= l && r <= R) return S[o][0];
pushdown(o);
int mid = (l + r) >> 1;
if (R <= mid) return Query(o << 1, l, mid, L, R);
if (L > mid) return Query(o << 1 | 1, mid + 1, r, L, R);
return Query(o << 1, l, mid, L, R) + Query(o << 1 | 1, mid + 1, r, L, R);
}
} seg;
int main() {
freopen("rps.in", "r", stdin);
freopen("rps.out", "w", stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> q;
cin >> str; str = ' ' + str;
seg.Build(1, 1, n);
while (q--) {
int op, l, r;
char x, y;
cin >> op >> l >> r;
if (op == 0) {
cin >> x >> y; x -= 'A', y -= 'A'; tmp[0] = p[0]; swap(tmp[0][(int)x], tmp[0][(int)y]);
seg.Change(1, 1, n, l, r, tmp[0]);
} else cin >> x, cout << (char)(seg.Query(1, 1, n, l, r)[x - 'A'] + 'A') << "\n";
}
return 0;
}
T3
蝴蝶图
设 \(G = (V, E)\) 为 \(L \cap R\) 的导出子图，那么我们考虑从 \(E\) 中的边都不连的时候往里加边。考虑 \(L\) 侧 MST 的变化。当我们加入 \(E\) 中的一些边，显然每次会删掉原来树上的一些边。那么这个时候抛开维护不谈，我们就应该注意到只有原 MST 中的边是有用的。然后再考虑有哪些边无论 \(V\) 中的联通状态如何都是一定在 MST 中的。发现如果将 \(E\) 中的边全连起来，此时还在 \(L\) 中的 MST 中的边无论如何一定在 MST 中。于是我们只需要考虑那些在第一棵 MST 中而不在第二棵 MST 中的边即可。这样的边只有 \(\mathcal{O}(|V|)\) 条，于是可以直接每次枚举贝尔数然后暴力对这些边重跑 MST。总复杂度 \(\mathcal{O}(Bell(|V|)V\alpha(|V|) + m\log m)\)，由于 \(|V|\) 很小，只有 11，因此可以通过。
代码
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
#include <map>
#define int long long
using namespace std;
const int inf = 0x3f3f3f3f3f3f3f;
int n, m, L, R, ans = inf;
map<int, int> mp[100005];
struct Edge { int u, v, w; } e[200005];
vector<Edge> el, er;
vector<int> cap;
int bel[200005], p[200005];
int dsu[100005], td[100005];
void clear(int x) { for (; x; --x) dsu[x] = x; }
int getf(int x) { return dsu[x] == x ? x : (dsu[x] = getf(dsu[x])); }
int f[100005];
int cur[15];
struct rD {
int f[100005], sz[100005];
vector<pair<int, int> > s;
void init(int x) { for (; x; --x) f[x] = x, sz[x] = 1; }
int getf(int x) { return f[x] == x ? x : getf(f[x]); }
bool Merge(int x, int y) {
x = getf(x), y = getf(y); (sz[x] > sz[y]) ? swap(x, y) : void();
return (x == y) ? 0 : (f[x] = y, sz[y] += sz[x], s.emplace_back(x, y), 1);
}
void Re(int t) {
while ((int)s.size() > t) {
int x, y; pair<int, int> p = s.back(); s.pop_back(); x = p.first, y = p.second;
f[x] = x, sz[y] -= sz[x];
}
}
} dl, dr;
int rl, rr;
void work(int t) {
int tl = dl.s.size(), tr = dr.s.size(), res = 0;
for (int i = 1; i <= t; i++) {
res += f[cur[i]];
for (int j = cur[i]; j; j -= (j & (-j))) {
dl.Merge(cap[63 - __builtin_clzll(j & (-j))], cap[63 - __builtin_clzll(cur[i] & (-cur[i]))]);
dr.Merge(cap[63 - __builtin_clzll(j & (-j))], cap[63 - __builtin_clzll(cur[i] & (-cur[i]))]);
}
}
if (res >= inf) return dl.Re(tl), dr.Re(tr);
for (auto E : el) {
int u = E.u, v = E.v;
if (dl.getf(u) != dl.getf(v)) dl.Merge(u, v), res += E.w;
}
for (auto E : er) {
int u = E.u, v = E.v;
if (dr.getf(u) != dr.getf(v)) dr.Merge(u, v), res += E.w;
}
dl.Re(tl), dr.Re(tr);
ans = min(ans, res);
}
void dfs(int x, int y) {
if (x == (int)cap.size()) return work(y);
for (int i = 1; i <= y; i++) {
cur[i] ^= (1 << x);
dfs(x + 1, y);
cur[i] ^= (1 << x);
}
cur[++y] = (1 << x);
dfs(x + 1, y);
}
int _f[200005];
signed main() {
freopen("butterfly.in", "r", stdin);
freopen("butterfly.out", "w", stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> L >> R;
for (int i = 1; i <= m; i++) {
int u, v, ww, x;
cin >> u >> v >> ww;
if (u == v) continue;
if (u > v) swap(u, v);
x = mp[u][v];
if (!x) mp[u][v] = ww;
else mp[u][v] = min(x, ww);
}
for (int i = 1; i <= L; i++) cin >> m, bel[m] |= 1;
for (int i = 1; i <= R; i++) cin >> m, bel[m] |= 2, (bel[m] == 3 ? cap.emplace_back(m) : void());
m = 0;
for (int i = 1; i <= n; i++) for (auto v : mp[i]) e[++m] = { i, v.first, v.second };
sort(e + 1, e + m + 1, [](Edge x, Edge y) { return x.w < y.w; });
for (int S = 0; S < (1 << (int)cap.size()); S++) {
for (int i = 0; i < (int)cap.size(); i++) p[dsu[cap[i]] = cap[i]] = ((S >> i) & 1);
int tmp = __builtin_popcountll(S);
for (int i = 1; i <= m; i++) {
int u = e[i].u, v = e[i].v;
if (bel[u] == 3 && bel[v] == 3 && p[u] && p[v] && getf(u) != getf(v)) dsu[dsu[u]] = v, --tmp, f[S] += e[i].w;
}
if (tmp != 1) f[S] = inf;
}
clear(n);
for (int i = 1; i <= m; i++) {
int u = e[i].u, v = e[i].v;
if (bel[u] == 3 && bel[v] == 3 && getf(u) != getf(v)) dsu[dsu[u]] = v;
}
for (int i = 1; i <= n; i++) td[i] = dsu[i];
for (int i = 1; i <= m; i++) {
int u = e[i].u, v = e[i].v;
if (((bel[u] == 1 && bel[v] == 1) || (bel[u] ^ bel[v]) == 2) && getf(u) != getf(v)) dsu[dsu[u]] = v, _f[i] = 2;
}
for (int i = 1; i <= n; i++) dsu[i] = td[i];
for (int i = 1; i <= m; i++) {
int u = e[i].u, v = e[i].v;
if (((bel[u] == 2 && bel[v] == 2) || (bel[u] ^ bel[v]) == 1) && getf(u) != getf(v)) dsu[dsu[u]] = v, _f[i] = 2;
}
clear(n);
for (int i = 1; i <= m; i++) {
int u = e[i].u, v = e[i].v;
if (((bel[u] == 1 && bel[v] == 1) || (bel[u] ^ bel[v]) == 2) && getf(u) != getf(v)) dsu[dsu[u]] = v, _f[i] |= 1;
}
clear(n);
for (int i = 1; i <= m; i++) {
int u = e[i].u, v = e[i].v;
if (((bel[u] == 2 && bel[v] == 2) || (bel[u] ^ bel[v]) == 1) && getf(u) != getf(v)) dsu[dsu[u]] = v, _f[i] |= 1;
}
dl.init(n), dr.init(n);
int _ = 0;
for (int i = 1; i <= m; i++) {
int u = e[i].u, v = e[i].v, t;
if (bel[u] == 3 && bel[v] == 3) continue;
if (_f[i] == 0) continue;
else if (_f[i] == 1) (bel[u] == 1 || bel[v] == 1) ? el.emplace_back(e[i]) : er.emplace_back(e[i]);
else t = ((bel[u] == 1 || bel[v] == 1) ? dl.Merge(u, v) : dr.Merge(u, v)), _ += t * e[i].w;
}
sort(el.begin(), el.end(), [](Edge x, Edge y) { return x.w < y.w; });
sort(er.begin(), er.end(), [](Edge x, Edge y) { return x.w < y.w; });
cur[1] = 1;
dfs(1, 1);
cout << _ + ans << "\n";
return 0;
}
T4
游戏厅
观察数据范围，一眼认出网络流。考虑 \(f_i\) 表示第 \(i\) 个人是否使用 A 机器。那么先建出时间轴，每个时间段上我们限制用 \(A\) 机器的人数。原本的上界是 \(x\)，但是由于 \(n\) 个人的存在，每个位置的上界可能就变了。而对于下界，设这个时间总共有 \(t_i\) 人，那么 \(B\) 机器的个数限制了用 \(A\) 的人数下界是 \(\max\{ 0, t_i - y \}\)。于是对 \(m\) 个人的每个，从 \(t_i\) 向 \(s_i\) 连边，无脑无源汇上下界可行流即可。构造方案是容易的。
代码
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
using namespace std;
const int inf = 0x3f3f3f3f;
int n, m, X, Y;
pair<int, int> p[605];
int d[1205], dcnt;
int d1[1205], d2[1205];
int lb[1205], rb[1205];
struct Workable_Flow_with_LRBounds {
int n, S, T, c;
int head[5005], cur[5005], nxt[100005], to[100005], cst[100005], res[100005], ecnt;
int add(int u, int v, int x) {
to[++ecnt] = v, nxt[ecnt] = head[u], head[u] = ecnt, res[ecnt] = x;
to[++ecnt] = u, nxt[ecnt] = head[v], head[v] = ecnt, res[ecnt] = 0;
return ecnt - 1;
}
int in[5005], out[5005];
int add(int u, int v, int l, int r) {
in[v] += l, out[u] += l;
return add(u, v, r - l);
}
queue<int> q;
int dep[5005];
bool bfs(int s, int t) {
for (int i = 1; i <= n + 2; i++) dep[i] = 0;
q.push(s);
dep[s] = 1;
while (!q.empty()) {
int x = q.front();
q.pop();
for (int i = head[x]; i; i = nxt[i]) {
int v = to[i];
if (!dep[v] && res[i] > 0) {
dep[v] = dep[x] + 1;
q.push(v);
}
}
}
return (dep[t] > 0);
}
int dfs(int x, int flow) {
if (x == T || !flow)
return flow;
int ret = 0;
for (int i = cur[x]; i && flow; i = nxt[i]) {
cur[x] = i;
int v = to[i];
if (dep[v] == dep[x] + 1 && res[i] > 0) {
int tmp = dfs(v, min(flow, res[i]));
res[i] -= tmp;
res[i ^ 1] += tmp;
ret += tmp;
flow -= tmp;
}
}
if (!ret)
dep[x] = 0;
return ret;
}
int dinic() {
int ret = 0;
while (bfs(S, T)) {
for (int i = 1; i <= T; i++) cur[i] = head[i];
ret += dfs(S, inf);
}
return ret;
}
void initialize(int nn) {
n = nn, ecnt = 1;
for (int i = 0; i <= n + 2; i++) in[i] = out[i] = head[i] = 0;
}
int work() {
S = n + 1, T = n + 2;
for (int i = 1; i <= n; i++) {
if (in[i] < out[i]) add(i, T, out[i] - in[i]);
else add(S, i, in[i] - out[i]);
}
int t = dinic();
for (int i = head[S]; i; i = nxt[i]) if (res[i]) return -1;
return t;
}
} G;
int eid[5005];
int f[5005];
int o[605], ans[605];
int ed[605];
int main() {
freopen("round.in", "r", stdin);
freopen("round.out", "w", stdout);
int tc;
cin >> tc;
while (tc--) {
memset(f, 0, sizeof f);
memset(ed, 0, sizeof ed);
memset(d1, 0, sizeof d1);
memset(d2, 0, sizeof d2);
memset(ans, 0, sizeof ans);
cin >> n >> m >> X >> Y; dcnt = 0;
for (int i = 1; i <= n + m; i++) cin >> p[i].first >> p[i].second, d[++dcnt] = p[i].first, d[++dcnt] = p[i].second, o[i] = i;
sort(d + 1, d + dcnt + 1);
dcnt = unique(d + 1, d + dcnt + 1) - d - 1;
for (int i = 1; i <= n + m; i++) {
int l = p[i].first = lower_bound(d + 1, d + dcnt + 1, p[i].first) - d;
int r = p[i].second = lower_bound(d + 1, d + dcnt + 1, p[i].second) - d;
++d1[l], --d1[r];
if (i <= n) ++d2[l], --d2[r];
}
for (int i = 1; i <= dcnt; i++) d1[i] += d1[i - 1], d2[i] += d2[i - 1];
bool no = 0;
G.initialize(dcnt);
for (int i = 1; i < dcnt; i++) {
lb[i] = max(0, d1[i] - d2[i] - Y), rb[i] = X - d2[i];
if (lb[i] > rb[i]) {
no = 1;
break;
}
G.add(i, i + 1, lb[i], rb[i]);
}
if (no) {
cout << "NO\n";
continue;
}
for (int i = n + 1; i <= n + m; i++) eid[i] = G.add(p[i].second, p[i].first, 0, 1);
if (G.work() == -1) {
cout << "NO\n";
continue;
}
for (int i = n + 1; i <= n + m; i++) f[i] = G.res[eid[i]];
sort(o + 1, o + n + m + 1, [](int x, int y) { return p[x] < p[y]; });
for (int i = 1; i <= n + m; i++) {
int t = o[i];
if (f[t]) {
for (int j = X + 1; j <= X + Y; j++) {
if (ed[j] <= p[t].first) {
ans[t] = j;
ed[j] = p[t].second;
break;
}
}
} else {
for (int j = 1; j <= X; j++) {
if (ed[j] <= p[t].first) {
ans[t] = j;
ed[j] = p[t].second;
break;
}
}
}
}
cout << "YES\n";
for (int i = 1; i <= n; i++) cout << ans[i] << " ";
cout << "\n";
for (int i = n + 1; i <= n + m; i++) cout << ans[i] << " ";
cout << "\n";
}
return 0;
}
贪心一定要构造出最优方案。
T3。
也可以有结论：将边集 \(E\) 分为 \(E_1, E_2\) 分别做 MST，得到的边集合并再做 MST 即为原图 MST。
T4。对时间考虑，而不是对人和机器考虑。不会做时考虑思考对象。
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