poj 1149（最大流）

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10400		Accepted: 4540

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

Source

Croatia OI 2002 Final Exam - First day

题目： http://poj.org/problem?id=1149

分析：这题一开始打算用猪圈作为点入手，然后用分层图来做，但是这样的点太多，还有复杂度太高，就没去尝试了，有谁这样过的和我分享下哈。。。

转个角度来分析，把每个卖猪的人作为点，当前来买猪的人如果先前买猪的人打开相同的猪圈，则先前那个人可以把所有能开的猪圈里的猪留给它，也就是在他们之间连一条无穷大的边，然后，如果是第一个开某个猪圈的人，就在源点与某人直接连上容量为猪圈中猪的数量的边，每个人与汇连上一条容量为它想买猪的数量的边，此时求最大流，最大流即答案。

描述是在是挫阿～～～看代码好理解

代码：

#include<cstdio>
using namespace std;
const int mm=2222222;
const int mn=2222;
const int oo=1000000000;
int node,src,dest,edge;
int ver[mm],flow[mm],next[mm];
int head[mn],work[mn],dis[mn],q[mn],num[mn];
int last[11111];
inline int min(int a,int b)
{
    return a<b?a:b;
}
inline void prepare(int _node,int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0;i<node;++i)head[i]=-1;
    edge=0;
}
inline void addedge(int u,int v,int c)
{
    ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0;i<node;++i)dis[i]=-1;
    dis[q[r++]=src]=0;
    for(l=0;l<r;++l)
        for(i=head[u=q[l]];i>=0;i=next[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)return 1;
            }
    return 0;
}
int Dinic_dfs(int u,int exp)
{
    if(u==dest)return exp;
    for(int &i=work[u],v,tmp;i>=0;i=next[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            return tmp;
        }
    return 0;
}
int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0;i<node;++i)work[i]=head[i];
        while(delta=Dinic_dfs(src,oo))ret+=delta;
    }
    return ret;
}
int main()
{
    int i,j,k,n,m;
    while(scanf("%d%d",&m,&n)!=-1)
    {
        for(i=1;i<=m;++i)last[i]=0,scanf("%d",&num[i]);
        prepare(n+2,0,n+1);
        for(i=1;i<=n;++i)
        {
            scanf("%d",&k);
            while(k--)
            {
                scanf("%d",&j);
                if(!last[j])addedge(src,i,num[j]);
                else addedge(last[j],i,oo);
                last[j]=i;
            }
            scanf("%d",&k);
            addedge(i,dest,k);
        }
        printf("%d\n",Dinic_flow());
    }
    return 0;
}