poj 1149 PIGS (最大流)

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11736		Accepted: 5184

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 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6

Sample Output

7

最大流的题
建图是关键
刚开始一直想把笼子算在图里面，各种不行
最后还是
参考大神的解题报告
http://hi.baidu.com/_%B3%D4%B8%D6%C7%D9/blog/item/9e43e235d2b3a035d5074254.html

#include<iostream>
#include<queue>
using namespace std;

const int INF = INT_MAX;

int m, n;
int pigs[1010];
int opend[1010];
int map[110][110];
int pre[1010];
int flow[1010];
int mmax;

int dfs()
{
    memset(pre, -1, sizeof(pre));
    memset(flow, 0, sizeof(flow));
    queue<int> q;
    pre[0]=0;
    flow[0]=INF;
    q.push(0);
    while(!q.empty())
    {
        int k=q.front();
        q.pop();
        if(k==n+1)
            break;
        for(int i=1;i<=n+1;i++)
        {
            if(pre[i]==-1 && map[k][i]!=0)
            {
                if(map[k][i]<flow[k])
                    flow[i]=map[k][i];
                else
                    flow[i]=flow[k];
                pre[i]=k;
                q.push(i);
            }
        }
    }
    if(pre[n+1]==-1)
        return -1;
    return flow[n+1];
}

int ek()
{
    mmax=0;
    int now, last;
    int res;
    while((res=dfs()) && res!=-1)
    {
        mmax+=res;
        now = n+1;
        while(now!=0)
        {
            last = pre[now];
            map[now][last]+=res;
            map[last][now]-=res;
            now = last;
        }        
    }
    return mmax;
}

int main()
{
    freopen("e:\\data.txt", "r", stdin);   
    freopen("e:\\out.txt", "w", stdout);
    while(cin>>m>>n)
    {
        memset(pigs, 0, sizeof(pigs));
        memset(opend, 0, sizeof(opend));
        for(int i=1;i<=m;i++)
        {
            cin>>pigs[i];
        }
        for(int i=1;i<=n;i++)
        {
            int k;
            cin>>k;
            while(k--)
            {
                int a;
                cin>>a;
                if(opend[a]==0)
                {
                    map[0][i]+=pigs[a];
                    opend[a]=i;
                }
                else
                {
                    map[opend[a]][i]=INF;
                    opend[a]=i;
                }
            }
            int b;
            cin>>b;
            map[i][n+1]=b;
        }
        cout<<ek()<<endl;
    }
    return 0;
}