POJ 1149 PIGS 最大流建模

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PIGS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 14204 Accepted: 6305

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

Source

Croatia OI 2002 Final Exam - First day

 

题意：有M个猪圈，每个猪圈里初始时有若干头猪。一开始所有猪圈都是关闭的。依次来了N个顾客，每个顾客分别会打开指定的几个猪圈，从中买若干头猪。每个顾客分别都有他能够买的数量的上限。每个顾客走后，他打开的那些猪圈中的猪，都可以被任意地调换到其它开着的猪圈里，然后所有猪圈重新关上。问总共最多能卖出多少头猪。（1 <= N <= 100, 1 <= M <= 1000）。

建图：

每个顾客分别用一个结点来表示。
对于每个猪圈的第一个顾客，从源点向他连一条边，容量就是该猪圈里的猪的初始数量。如果从源点到一名顾客有多条边，则可以把它们合并成一条，容量相加。
 对于每个猪圈，假设有n个顾客打开过它，则对所有整数i∈[1, n)，从该猪圈的第i个顾客向第i + 1个顾客连一条边，容量为∞。
 从各个顾客到汇点各有一条边，容量是各个顾客能买的数量上限。

 

 

#include<stdio.h>#include<string.h>#include<queue>using namespace std;#define INF 9999999int n,m;int map[300][300],pre[300],flow[300][300],p[300],a[300];int EK(int s,int t){    int sum=0;    queue<int>q;    memset(flow,0,sizeof(flow));    for(;;)    {        memset(a,0,sizeof(a));//记录残量        a[s]=INF;        q.push(s);        while(!q.empty())        {            int u=q.front();            q.pop();            for(int i=1; i<=t; i++)                if(!a[i]&&map[u][i]>flow[u][i])                {                    p[i]=u;//记录i的父亲节的是u                    q.push(i);                    a[i]=a[u]<map[u][i]-flow[u][i]?a[u]:map[u][i]-flow[u][i];                }        }        if(!a[t])break;//如果残量是0的话，就找到最大流        for(int i=t; i!=s; i=p[i])//每条路加上最小残量        {            flow[p[i]][i]+=a[t];            flow[i][p[i]]-=a[t];        }        sum+=a[t];//记录流量    }    return sum;}int main(){    int a,b,c,max;    int pig[1007],link[1007];    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(map,0,sizeof(map));        memset(p,0,sizeof(p));        memset(link,0,sizeof(link));        memset(pig,0,sizeof(pig));        int t=m+1;        for(int i=1;i<=n;i++)            scanf("%d",&pig[i]);        for(int i=1;i<=m;i++)        {            scanf("%d",&a);            for(int j=1;j<=a;j++)            {                scanf("%d",&b);                if(link[b]==0)                {                    link[b]=i;                    map[0][i]+=pig[b];                }                else                {                    map[link[b]][i]=INF;                }            }            scanf("%d",&map[i][t]);        }        max=EK(0,t);        printf("%d\n",max);    }    return 0;}