poj1149 最大流 PIGS

题意：M个猪圈，N个顾客，每个顾客有一些的猪圈的钥匙，只能购买这些有钥匙的猪圈里的猪，而且要买一定数量的猪，每个猪圈有已知数量的猪，
但是猪圈可以重新打开，将猪的个数，重新分配，以达到卖出的猪的数量最多。

思路：刚学网络流，表示很菜很菜很菜~~
①构造网络，将顾客看成源点和汇点以外的结点，并设另外两个节点：源点和汇点。
②源点和每个猪圈的第一个顾客连边，边的权是开始时候猪圈中猪的数量。
③ 若源点和某个节点之间有重边，则将权合并
④顾客j紧跟顾客i之后打开某个猪圈，则<i.j>的权是正无穷。
⑤每个顾客和会点之间连边，边的权值是顾客所希望购买的猪的数量。
例如：样例中的就可以建立如图：

其中inf是正无穷~~

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 33 1 102 1 2 22 1 3 31 2 6

Sample Output

7

#include<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<cstring>using namespace std;#define maxn 1005#define INF 0x3f3f3f3fint cap[maxn][maxn],flow[maxn][maxn];int a[maxn],p[maxn];int n,m;int EK(){    int s=0,t=n+1;    queue<int>q;    int f=0;    for(;;)    {        memset(a,0,sizeof(a));        a[s]=INF;        q.push(s);        while(!q.empty())        {            int u=q.front();            q.pop();            for(int v = 0; v <=t; v++)                if(!a[v] && cap[u][v] > flow[u][v])                {                    p[v] = u;                     q.push(v);                    a[v] = min(a[u], cap[u][v]-flow[u][v]);                }        }        if(a[t]==0)            break;        for(int u=t; u!=s; u=p[u])        {            flow[p[u]][u]+=a[t];            flow[u][p[u]]-=a[t];        }        f+=a[t];    }    return f;}int main(){    int pig[maxn]={0},vist[maxn]={0};    int k,key;    memset(cap,0,sizeof(cap));    scanf("%d %d",&m,&n);    for(int i=1;i<=m;i++)        scanf("%d",&pig[i]);    for(int i=1;i<=n;i++)    {        scanf("%d",&k);        for(int j=1;j<=k;j++)            {                scanf("%d",&key);        if(!vist[key])            cap[0][i]+=pig[key];            else                cap[vist[key]][i]=INF;            vist[key]=i;        }        scanf("%d",&key);        cap[i][n+1]=key;    }    printf("%d\n",EK());    return 0;}