3 1
1 10
2 10
10 3
2 3 1000

13

最小割的模板，事实上就是最大流

注意：

Verson 2:

1.修复Bug。在本次模板中改动了q队列的长度。

题目。裸最小割，

设S为用模块A的集合，T为模块B

则

s->任务i  //模块B的cost

任务i->t  //模块A的cost

任务i->任务j //(i,j)不在一个集合的cost。注意最小割，要2个方向（最小割仅仅保证s->t没路径，t->s的路径不用割）

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXn (20000+10)
#define MAXm (200000+10)
#define MAXN (MAXn+2)
#define MAXM ((MAXn*2+MAXm*2)*2+100)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
class Max_flow  //dinic+当前弧优化   
{    
public:    
    int n,s,t;    
    int q[MAXN];    
    int edge[MAXM],next[MAXM],pre[MAXN],weight[MAXM],size;    
    void addedge(int u,int v,int w)      
    {      
        edge[++size]=v;      
        weight[size]=w;      
        next[size]=pre[u];      
        pre[u]=size;      
    }      
    void addedge2(int u,int v,int w){addedge(u,v,w),addedge(v,u,0);}     
    bool b[MAXN];    
    int d[MAXN];    
    bool SPFA(int s,int t)      
    {      
        For(i,n) d[i]=INF;    
        MEM(b)    
        d[q[1]=s]=0;b[s]=1;      
        int head=1,tail=1;      
        while (head<=tail)      
        {      
            int now=q[head++];      
            Forp(now)      
            {      
                int &v=edge[p];      
                if (weight[p]&&!b[v])      
                {      
                    d[v]=d[now]+1;      
                    b[v]=1,q[++tail]=v;      
                }      
            }          
        }      
        return b[t];      
    }     
    int iter[MAXN];  
    int dfs(int x,int f)  
    {  
        if (x==t) return f;  
        Forpiter(x)  
        {  
            int v=edge[p];  
            if (weight[p]&&d[x]<d[v])  
            {  
                  int nowflow=dfs(v,min(weight[p],f));  
                  if (nowflow)  
                  {  
                    weight[p]-=nowflow;  
                    weight[p^1]+=nowflow;  
                    return nowflow;  
                  }  
            }  
        }  
        return 0;  
    }  
    int max_flow(int s,int t)  
    {  
        int flow=0;  
        while(SPFA(s,t))  
        {  
            For(i,n) iter[i]=pre[i];  
            int f;  
            while (f=dfs(s,INF))  
                flow+=f;   
        }  
        return flow;  
    }   
    void mem(int n,int s,int t)    
    {    
        (*this).n=n;  
        (*this).t=t;    
        (*this).s=s;    
          
        size=1;    
        MEM(pre)   
    }    
}S;    

int n,m;

int main()
{
//	freopen("poj3469.in","r",stdin);
//	freopen(".out","w",stdout);
	
	scanf("%d%d",&n,&m);
	
	int s=1,t=n+2;
	S.mem(n+2,s,t); 
	For(i,n)
	{
		int ai,bi;
		scanf("%d%d",&ai,&bi);
		S.addedge2(i+1,t,ai);
		S.addedge2(s,i+1,bi);
	}
	
	For(i,m)
	{
		int a,b,w;
		scanf("%d%d%d",&a,&b,&w);
		S.addedge(a+1,b+1,w);
		S.addedge(b+1,a+1,w);
	}
	
	cout<<S.max_flow(s,t)<<endl;
	
	
	return 0;
}