bzoj 2527: [Poi2011]Meteors

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;
			else
				erfen[id].first = i + 1;
			++js;
		}
	}
	solve();
}
int main() {
	//read
	n = 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;
	//work
	solve();
	//print
	for(int i = 1;i <= n; ++i) {
		if(ans[i]) printf("%d
", ans[i]);
		else puts("NIE");
	}
	return 0;
}