Dinic问题

问题：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

回答：题目大意是第i个moduel在两个核的运行时间分别为ai、bi，两个modue之间的数据交换要有花费。可以将其建成一个网络流模型，源点和汇点分别是两个核，记为0，n+1，然后建图，求最小割即可。

#include<cstdio>  
    #include<cstring>  
    #include<iostream>  
    using namespace std;  
      
    const int N=20100;  
    const int M=200200;  
    const int inf=1<<29;  
    struct node  
    {  
     int v,f;  
     int next;  
    }edge[8*M];  
    int head[N],num;  
    int n,m;  
    int s,t,NN;  
      
    void init()  
    {  
     for(int i=0;i<=n+2;i++)  
      head[i]=-1;  
     num=0;  
    }  
      
    void addege(int u,int v,int f)  
    {  
     edge[num].v=v;  
     edge[num].f=f;  
     edge[num].next=head[u];  
     head[u]=num++;  
     edge[num].v=u;  
     edge[num].f=0;  
     edge[num].next=head[v];  
     head[v]=num++;  
    }  
      
    int sap()  
    {  
     int pre[N],cur[N],dis[N],gap[N];  
     int flow=0,aug=inf,u;  
     bool flag;  
     int i;  
     for(i=1;i<=NN;i++)  
     {  
      cur[i]=head[i];  
      gap[i]=dis[i]=0;  
     }  
     gap[s]=NN;  
     u=pre[s]=s;  
     while(dis[s]<NN)  
     {  
      flag=0;  
      for(int &j=cur[u];j!=-1;j=edge[j].next)  
      {  
       int v=edge[j].v;  
       if(edge[j].f>0 && dis[u]==dis[v]+1)  
       {  
        flag=1;  
        if(edge[j].f<aug) aug=edge[j].f;  
        pre[v]=u;  
        u=v;  
        if(u==t)  
        {  
         flow+=aug;  
         while(u!=s)  
         {  
          u=pre[u];  
          edge[cur[u]].f-=aug;  
          edge[cur[u]^1].f+=aug;  
         }  
         aug=inf;  
        }  
        break;  
       }  
      }  
      if(flag) continue;  
      int mindis=NN;  
      for(int j=head[u];j!=-1;j=edge[j].next)  
      {  
       int v=edge[j].v;  
       if(edge[j].f>0 && dis[v]<mindis)  
       {  
        mindis=dis[v];  
        cur[u]=j;  
       }  
      }  
      if((–gap[dis[u]])==0) break;  
      gap[dis[u]=mindis+1]++;  
      u=pre[u];  
     }  
     return flow;  
    }  
      
    int main()  
    {  
     scanf(“%d%d”,&n,&m);  
     int i,j;  
     int a,b,w;  
     s=0;t=n+1;NN=n+2;  
     init();  
     for(i=1;i<=n;i++)  
     {  
      scanf(“%d%d”,&a,&b);  
      addege(s,i,a);  
      addege(i,t,b);  
     }  
     for(i=1;i<=m;i++)  
     {  
      scanf(“%d%d%d”,&a,&b,&w);  
      addege(a,b,w);  
      addege(b,a,w);  
     }  
     printf(“%d/n”,sap());  
     return 0;  
    }