poj 1149 PIGS(网络流 BFS 标号法)(困难)

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18402		Accepted: 8373

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

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define INF 300000000
#define MAXM 1000
#define MAXN 100
int s,t;//源点，汇点
int cus[MAXN+2][MAXN+2];//N+2个结点之间的容量
int flow[MAXN+2][MAXN+2];//结点之间的流量
int i,j;

void init()
{
    int M,N;
    int num;
    int k;
    int house[MAXN];
    int last[MAXN];
    memset(last,0,sizeof(last));memset(cus,0,sizeof(cus));
    scanf("%d%d",&M,&N);
    s=0,t=N+1;
    for(i=1;i<=M;i++)
        scanf("%d",&house[i]);
    for(i=1;i<=N;i++)
    {
        scanf("%d",&num);
        for(j=0;j<num;j++)
        {
            scanf("%d",&k);
            if(last[k]==0)
                cus[s][i]=cus[s][i]+house[k];
            else
                cus[last[k]][i]=INF;
            last[k]=i;
        }
        scanf("%d",&cus[i][t]);
    }
}

void ford()
{
    int pre[MAXN+2];
    int minflow[MAXN+2];
    int queue[MAXN+2];
    int qs,qe;
    int v;  //当前检查的顶点
    int p;
    for(i=0;i<MAXN+2;i++)
    {
        for(j=0;j<MAXN+2;j++)
            flow[i][j]=0;
    }
    minflow[0]=INF;
    while(1)
    {
        for(i=0;i<MAXN+2;i++)
        {
            pre[i]=-2;
        }
        pre[0]=-1;
        qs=0;queue[qs]=0;qe=1;
        while(qs<qe && pre[t]==-2)
        {
            v=queue[qs];qs++;
            for(i=0;i<t+1;i++)
            {
                if(pre[i]==-2 && (p=cus[v][i]-flow[v][i]))
                {
                    pre[i]=v;queue[qe]=i;qe++;
                    minflow[i]=(minflow[v]<p)?minflow[v]:p;
                }
            }
        }
        if(pre[t]==-2) break;
        for(i=pre[t],j=t;i!=-1;j=i,i=pre[i])
        {
            flow[i][j]=flow[i][j]+minflow[t];
            flow[j][i]=-flow[i][j];
        }
    }
    for(i=0,p=0;i<t;i++)
    {
        p=p+flow[i][t];
    }
    printf("%d\n",p);
}

int main()
{
    init();
    ford();
    return 0;
}