昨天写了一晚，越写复杂度越感觉不对，早上一想果然是假的。

(这里n,m,k我就不区分了)

首先一个城市的询问可以很容易的二分

check用树状数组维护区间(区间修改，单点查询的那种)

一次是\(O(nlog^2n)\)

n次就是\(O(n^2log^2n)\)

但是我们check的时候都是树状数组维护，询问相同

我们就可以整体二分(顾名思义)

把区间考虑成二叉树(类似线段树的形状)

我们每一层用一遍树状数组

查询的话，一个国家用一个链表存下所在的点

因为深度是\(logn\)的

复杂度是还是差不多的\((Onlog^2n)\)

妙啊

会炸long long 及时break就好

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <utility>
#include <vector>
#define ll long long
using namespace std;
const int N = 5e5+7;
int read() {int x = 0,f = 1;char s = getchar();for(; s > '9'||s < '0'; s = getchar()) if(s == '-') f = -1;for(; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';return x * f;
}
int n, m, k;
int L[N], R[N], w[N], need[N], ans[N];
pair<int,int> erfen[N];
vector<int> hav[N];
namespace BIT {ll sum[N];void add(int id, ll ad) {for(int i = id; i <= m; i += (i&-i)) sum[i] += ad;}ll query(int id) {ll ans = 0;for(int i = id; i >= 1;i -= (i&-i)) ans += sum[i]; return ans;}void modify(int l, int r, ll ad) {add(l, ad), add(r + 1, -ad);}void clear() {memset(sum, 0, sizeof(sum));}
}
void solve() {vector<pair<int,int> > Q;for(int i = 1; i <= n; ++i)if(erfen[i].first!=k+1&&erfen[i].first <= erfen[i].second)Q.push_back(make_pair((erfen[i].first+erfen[i].second)>>1,i));if(Q.empty()) return;sort(Q.begin(), Q.end());BIT::clear();for(int i = 1, js = 0;i <= k; ++ i) {if(L[i] <= R[i]) BIT::modify(L[i], R[i], (ll)w[i]);else BIT::modify(L[i], m, w[i]), BIT::modify(1, R[i], (ll)w[i]);while(js < Q.size() && Q[js].first == i) {int id = Q[js].second;ll tmp=0;for(vector<int>::iterator it = hav[id].begin();it != hav[id].end(); ++it) {tmp += BIT::query(*it);if(tmp>=(ll)need[id]) break;}if(tmp>=(ll)need[id]) ans[id] = i,erfen[id].second = i - 1;elseerfen[id].first = i + 1;++js;}}solve();
}
int main() {//readn = read(), m = read();for(int i = 1; i <= m; ++i) {int x = read();hav[x].push_back(i);}for(int i = 1; i <= n; ++i) need[i] = read();k = read();for(int i = 1; i <= k; ++i) L[i] = read(), R[i] = read(), w[i] = read();for(int i = 1; i <= n; ++i) erfen[i].first = 1, erfen[i].second = k+1;//worksolve();//printfor(int i = 1;i <= n; ++i) {if(ans[i]) printf("%d\n", ans[i]);else puts("NIE");}return 0;
}