poj 1149 PIGS（最大流）

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19617		Accepted: 8973

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

具体可见： 点击打开链接

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define N 1100
#define M 110
#define INF 0x3f3f3f3f
struct Edge
{
    int to,next,cap;
}edge[3*(N*M+2*M)];
int cnt;
int pig_num[N],vis[N];
int ma[M],head[N],d[N];
void init()
{
    memset(vis,0,sizeof(vis));
    memset(ma,0,sizeof(ma));
    memset(head,-1,sizeof(head));
    cnt=0;
}
void addedge(int from,int to,int cap)
{
    edge[cnt].to=to;edge[cnt].cap=cap;
    edge[cnt].next=head[from];head[from]=cnt++;

    edge[cnt].to=from;edge[cnt].cap=0;
    edge[cnt].next=head[to];head[to]=cnt++;
}

int bfs(int s,int t)
{
    memset(d,-1,sizeof(d));
    queue<int> q;
    q.push(s);
    d[s]=0;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(d[v]==-1&&edge[i].cap>0)
            {
                d[v]=d[u]+1;
                q.push(v);
            }
        }
    }
    return d[t]!=-1;
}
int dfs(int s,int t,int f)
{
    if(s==t||f==0) return f;
    int flow=0;
    for(int i=head[s];i!=-1&&flow<f;i=edge[i].next)
    {
        int v=edge[i].to;
        if(d[v]==d[s]+1&&edge[i].cap>0)
        {
            int x=min(f-flow,edge[i].cap);
            x=dfs(v,t,x);
            flow+=x;
            edge[i].cap-=x;
            edge[i^1].cap+=x;
            if(f==0) break;
        }
    }
    if(!flow) d[s]=-2;
    return flow;
}
int Dinic(int s,int t)///起点s终点t
{
    int flow=0,f;
    while(bfs(s,t))
    {
        while(f=dfs(s,t,INF))
            flow+=f;
    }
    return flow;
}
int main()
{
    int n,m,t,num;
    while(~scanf("%d %d",&n,&m))
    {
        int S=0,T=m+1;
        init();
        for(int i=1;i<=n;i++)
            scanf("%d",&pig_num[i]);
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&t);
            while(t--)
            {
                scanf("%d",&num);
                if(!vis[num])
                {
                    vis[num]=i;
                    ma[i]+=pig_num[num];
                }
                else addedge(vis[num],i,INF);
            }
            scanf("%d",&t);
            addedge(i,T,t);
        }
        for(int i=1;i<=m;i++)
            if(ma[i])
            addedge(S,i,ma[i]);
        int ans=Dinic(S,T);
        printf("%d\n",ans);
    }
    return 0;
}