poj 1149 PIGS 最大网络流

Description

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs. 
All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold. 
More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses. 
An unlimited number of pigs can be placed in every pig-house. 
Write a program that will find the maximum number of pigs that he can sell on that day.

Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N. 
The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000. 
The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line): 
A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

Output

The first and only line of the output should contain the number of sold pigs.

Sample Input

3 33 1 102 1 2 22 1 3 31 2 6

Sample Output

7

Source

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<deque>
using namespace std;
#define maxn 550
#define INF 99999999


int Map[maxn][maxn];
int prev[maxn];
bool vis[maxn];
int come[1100],pig[1100];


void EK(int x,int n)
{
    int sum=0;
    int temp,u,v,i;
    while(1)
    {
       memset(vis,false,sizeof(vis));
       memset(prev,0,sizeof(prev));
       deque<int>Q;
       prev[x]=0;
       vis[x]=true;
       Q.push_back(x);
       while(!Q.empty())
       {
       // printf("**********\n");
           u=Q.front();
           Q.pop_front();
           for(v=1;v<=n;v++)
           {
              //printf("+++++++++++\n");
               if(!vis[v]&&Map[u][v])
               {
                   prev[v]=u;
                   vis[v]=true;
                   if(v==n)
                   {
                       Q.clear();
                       break;
                   }
                   else
                    Q.push_back(v);
               }
           }
       }
       if(!vis[n])
        break;
       temp=INF;
       i=n;
       while(prev[i])
       {
           if(temp>Map[prev[i]][i])
                temp=Map[prev[i]][i];
            //printf("+++++++++++\n");
           i=prev[i];
       }
       i=n;
       while(prev[i])
       {
           Map[prev[i]][i]-=temp;
           Map[i][prev[i]]+=temp;
           // printf("**********\n");
           i=prev[i];
       }
       //printf("%d\n",temp);
       sum+=temp;
    }
    printf("%d\n",sum);
}


int main()
{
    int m,n;
    int i,j,k;
    int t;
    int x;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        k=n+1;
        memset(come,0,sizeof(come));
        memset(Map,0,sizeof(Map));
        for(i=1;i<=m;i++)
            scanf("%d",&pig[i]);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&t);
            for(j=1;j<=t;j++)
            {
                scanf("%d",&x);
                if(!come[x])
                    Map[k][i]+=pig[x];
                else
                    Map[come[x]][i]=INF;
                come[x]=i;//记录猪圈的指向
            }
            scanf("%d",&Map[i][k+1]);
        }
        EK(k,k+1);
    }
   return 0;
}