2018.06.27Dual Core CPU（最小割）

Dual Core CPU

Time Limit: 15000MS Memory Limit: 131072K

Total Submissions: 26136 Accepted: 11270

Case Time Limit: 5000MS

Description

As more and more computers are equipped with dual core CPU, SetagLilb, the Chief Technology Officer of TinySoft Corporation, decided to update their famous product - SWODNIW.

The routine consists of N modules, and each of them should run in a certain core. The costs for all the routines to execute on two cores has been estimated. Let’s define them as Ai and Bi. Meanwhile, M pairs of modules need to do some data-exchange. If they are running on the same core, then the cost of this action can be ignored. Otherwise, some extra cost are needed. You should arrange wisely to minimize the total cost.

Input

There are two integers in the first line of input data, N and M (1 ≤ N ≤ 20000, 1 ≤ M ≤ 200000) .

The next N lines, each contains two integer, Ai and Bi.

In the following M lines, each contains three integers: a, b, w. The meaning is that if module a and module b don’t execute on the same core, you should pay extra w dollars for the data-exchange between them.

Output

Output only one integer, the minimum total cost.

Sample Input

3 1

1 10

2 10

10 3

2 3 1000

Sample Output

13

Source

POJ Monthly–2007.11.25, Zhou Dong

由于要将nnn个点分成两组，那么这显然是“割”的意思，再仔细想想，如果我们建立源点sss和汇点ttt，从sss向每个点连容量为AiA_iAi​的边，从每个点向ttt连容量为BiB_iBi​的边，然后两个点i,ji,ji,j连容量为www的无向边，建好图后我们只用求出最小割即可。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#define N 250000
#define M 3000000
using namespace std;
inline long long read(){
	long long ans=0;
	char ch=getchar();
	while(!isdigit(ch))ch=getchar();
	while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar();
	return ans;
}
int n,m,s,t,ans=0,d[N],first[N],cnt=-1;
struct Node{int v,next,c;}e[M<<1];
inline void add(int u,int v,int c){
	e[++cnt].v=v;
	e[cnt].next=first[u];
	e[cnt].c=c;
	first[u]=cnt;
	e[++cnt].v=u;
	e[cnt].next=first[v];
	e[cnt].c=0;
	first[v]=cnt;
}
inline bool bfs(){
	queue<int>q;
	q.push(s);
	memset(d,-1,sizeof(d));
	d[s]=0;
	while(!q.empty()){
		int x=q.front();
		q.pop();
		for(int i=first[x];i!=-1;i=e[i].next){
			int v=e[i].v;
			if(d[v]!=-1||e[i].c<=0)continue;
			d[v]=d[x]+1;
			if(v==t)return true;
			q.push(v);
		}
	}
	return false;
}
inline int dfs(int x,int f){
	if(x==t||!f)return f;
	int flow=f;
	for(int i=first[x];i!=-1;i=e[i].next){
		int v=e[i].v;
		if(flow&&e[i].c>0&&d[v]==d[x]+1){
			int tmp=dfs(v,min(flow,e[i].c));
			if(!tmp)d[v]=-1;
			e[i].c-=tmp;
			e[i^1].c+=tmp;
			flow-=tmp;
		}
	}
	return f-flow;
}
int main(){
	memset(first,-1,sizeof(first));
	n=read(),m=read(),s=0,t=n+1;
	for(int i=1;i<=n;++i){
		int a=read(),b=read();
		add(s,i,a),add(i,t,b);
	}
	for(int i=1;i<=m;++i){
		int a=read(),b=read(),w=read();
		add(a,b,w),add(b,a,w);
	}
	while(bfs())ans+=dfs(s,0x3f3f3f3f);
	printf("%d",ans);
	return 0;
}