CF903G Yet Another Maxflow Problem

题面

一张图分为两部分，左右都有(n)个节点，
(A_i->A_{i+1})连边，(B_{i}->B_{i+1})连边，容量给出
有(m)对(A_i->B_j)有边，容量给出
两种操作
1.修改某条(A_i->A_{i+1})的边的容量
2.询问从(A_1)到(B_n)的最大流
(n,m<=100000),流量(<=10^9)

Sol

首先最大流等于最小割
而且每条(A_i->A_{i+1})的边选择的最小割除了它自己以外都不会变
处理出每条(A)的边所选的最小割，就变成了单点修改全局查询
然后怎么处理
枚举(A)，每次把(A)连出去的到(B)的所有边的边权加到这个(B)之前的所有边
即区间加法，因为割的话会割到连到后面的边

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e5 + 5);

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

int n, m, q, w1[_], w2[_];
ll mf[_], mn[_ << 2], tag[_ << 2];
vector <int> G[_], W[_];

IL void Build1(RG int x, RG int l, RG int r){
	if(l == r){
		mn[x] = w2[l];
		return;
	}
	RG int mid = (l + r) >> 1;
	Build1(x << 1, l, mid), Build1(x << 1 | 1, mid + 1, r);
	mn[x] = min(mn[x << 1], mn[x << 1 | 1]);
}

IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int v){
	if(L <= l && R >= r){
		mn[x] += v, tag[x] += v;
		return;
	}
	RG int mid = (l + r) >> 1;
	if(L <= mid) Modify(x << 1, l, mid, L, R, v);
	if(R > mid) Modify(x << 1 | 1, mid + 1, r, L, R, v);
	mn[x] = min(mn[x << 1], mn[x << 1 | 1]) + tag[x];
}

IL void Build2(RG int x, RG int l, RG int r){
	tag[x] = 0;
	if(l == r){
		mn[x] = w1[l] + mf[l];
		return;
	}
	RG int mid = (l + r) >> 1;
	Build2(x << 1, l, mid), Build2(x << 1 | 1, mid + 1, r);
	mn[x] = min(mn[x << 1], mn[x << 1 | 1]);
}

int main(RG int argc, RG char *argv[]){
	n = Input(), m = Input(), q = Input();
	for(RG int i = 1; i < n; ++i) w1[i] = Input(), w2[i] = Input();
	for(RG int i = 1; i <= m; ++i){
		RG int u = Input(), v = Input(), w = Input();
		G[u].push_back(v), W[u].push_back(w);
	}
	Build1(1, 0, n - 1);
	for(RG int i = 0, l = G[1].size(); i < l; ++i)
		Modify(1, 0, n - 1, 0, G[1][i] - 1, W[1][i]);
	for(RG int i = 1; i < n; ++i){
		mf[i] = mn[1];
		for(RG int j = 0, l = G[i + 1].size(); j < l; ++j)
			Modify(1, 0, n - 1, 0, G[i + 1][j] - 1, W[i + 1][j]);
	}
	mf[n] = mn[1], Build2(1, 1, n);
	printf("%lld
", mn[1]);
	for(RG int i = 1; i <= q; ++i){
		RG int p = Input(), w = Input();
		Modify(1, 1, n, p, p, w - w1[p]);
		w1[p] = w;
		printf("%lld
", mn[1]);
	}
	return 0;
}