轉化為維護最大連續子串和

#include<bits/stdc++.h> #define ll long long #define pi pair<int, ll> #define mk make_pair using namespace std; const int maxn = 2010; struct node {int x, y;ll w;bool operator<(const node& t) const {if (y == t.y)return x < t.x;return y > t.y;} }p[maxn]; int X[maxn], Y[maxn]; ll mx[maxn * 4], L[maxn * 4], R[maxn * 4], sum[maxn * 4]; vector<pi> G[maxn]; #define ls o * 2 #define rs o * 2 + 1 #define mid (l + r) / 2 void up(int o, int l, int r, int k, ll w) {if (l == r) {mx[o] += w;L[o] += w;R[o] += w;sum[o] += w;return;}if (k <= mid)up(ls, l, mid, k, w);elseup(rs, mid + 1, r, k, w);sum[o] = sum[ls] + sum[rs];L[o] = max(L[ls], sum[ls] + L[rs]);R[o] = max(R[rs], sum[rs] + R[ls]);mx[o] = max(L[o], R[o]);ll tmp = max(mx[ls], mx[rs]);ll tmp2 = max(tmp, R[ls] + L[rs]);mx[o] = max(mx[o], tmp2); } int main() {int T;scanf("%d", &T);while (T--) {int n, m = 0;scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d%d%lld", &p[i].x, &p[i].y, &p[i].w);X[i] = p[i].x;Y[i] = p[i].y;}sort(X + 1, X + 1 + n);sort(Y + 1, Y + 1 + n);int sz1 = unique(X + 1, X + 1 + n) - X - 1;int sz2 = unique(Y + 1, Y + 1 + n) - Y - 1;for (int i = 1; i <= n; i++) {p[i].x = lower_bound(X + 1, X + 1 + sz1, p[i].x) - X;p[i].y = lower_bound(Y + 1, Y + 1 + sz2, p[i].y) - Y;G[p[i].y].push_back(mk(p[i].x, p[i].w));}ll ans = 0;for (int i = 1; i <= sz2; i++) { //枚舉下邊界for (int j = 1; j <= sz1 * 4; j++)mx[j] = L[j] = R[j] = sum[j] = 0;for (int j = i; j; j--) { //枚舉上邊界for (auto tmp : G[j])up(1, 1, sz1, tmp.first, tmp.second);ans = max(ans, mx[1]);}}printf("%lld\n", ans);for (int i = 1; i <= sz2; i++)G[i].clear();} }

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