poj 1149 PIGS

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 arives, 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 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6

Sample Output

7

 

//

 

源点引出m条边分别代表每个猪圈猪的个数，如果顾客i是第一个开猪圈p的人，则源点向i连p的边，如果i是第k个打开p的人，则从第k-1个打开p的门的人向k引边，最后每个人向汇点连边。

这样一个流的含义就能够把猪圈之间的调换包含进去了

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int inf=(1<<30);
const int point_num=300;
int cap[point_num][point_num],dist[point_num],gap[point_num];//初始化见main里面
int s0,t0,n;//源,汇和点数
int find_path(int p,int limit=0x3f3f3f3f)
{
    if(p==t0)   return limit;
    for(int i=0;i<n;i++)
    if(dist[p]==dist[i]+1  &&  cap[p][i]>0)
    {
       int t=find_path(i,min(cap[p][i],limit));
       if(t<0)   return t;
       if(t>0)
       {
            cap[p][i]-=t;
            cap[i][p]+=t;
            return t;
       }
    }
    int label=n;
    for(int i=0;i<n;i++)  if(cap[p][i]>0)  label=min(label,dist[i]+1);
    if(--gap[dist[p]]==0  ||  dist[s0]>=n )   return -1;
    ++gap[dist[p]=label];
    return 0;
}
int sap()
{
    //初始化s,t
    s0=0,t0=n-1;
    int t=0,maxflow=0;
    gap[0]=n;
    while((t=find_path(s0))>=0) maxflow+=t;
    return maxflow;
}
int pig[1100];
int vis[1100];
int main()
{
    int m;
    while(scanf("%d%d",&m,&n)==2)
    {
        //初始化
        memset(cap,0,sizeof(cap));
        memset(dist,0,sizeof(dist));
        memset(gap,0,sizeof(gap));
        //初始化cap
        for(int i=1;i<=m;i++) scanf("%d",&pig[i]);
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            int k;scanf("%d",&k);
            for(int j=0;j<k;j++)
            {
                int x;scanf("%d",&x);
                if(vis[x]==0)
                {
                    cap[0][i]+=pig[x];
                }
                else
                {
                    cap[vis[x]][i]=inf;
                }
                vis[x]=i;
            }
            scanf("%d",&cap[i][n+1]);
        }
        n+=2;
        printf("%d\n",sap());
    }
    return 0;
}