POJ-1149-PIGS（最大流 标号法）

PIGS
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 20033 Accepted: 9179
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<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;//标号法求矩阵网络最大流const int maxn_N=105;//顶点const int maxn_M=1005;//弧const int INF=0x3f3f3f3f;int customer[maxn_N][maxn_N];//容量int flow[maxn_M][maxn_M];//流量int s,t;//源点和汇点void init()//构图{    int M;//猪圈数量    int N;//顾客数量    int house[maxn_M];//初始数    int last[maxn_M];//回溯存路径    memset(last,0,sizeof(last));    memset(customer,0,sizeof(customer));    scanf("%d%d",&M,&N);    s=0;//源点    t=N+1;//汇点    for(int i=1; i<=M; i++)        scanf("%d",&house[i]);    for(int i=1; i<=N; i++)    {        int num;//每个顾客拥有的钥匙数量        scanf("%d",&num);        for(int j=0; j<num; j++)        {            int k;//钥匙序号            scanf("%d",&k);            last[k]==0?customer[s][i]+=house[k]:customer[last[k]][i]=INF;            last[k]=i;        }        scanf("%d",&customer[i][t]);//顾客到汇点，权值为顾客买猪数量    }//    printf("*****\n");}void ford()//标号法{    int prev[maxn_N];//回溯记录路径    int min_flow[maxn_N];//可改进量    int queue[maxn_M];//数组模拟队列    int q_star,q_end;//队列两头指针    int p;//cij-fij    memset(flow,0,sizeof(flow));//初始流量为零流    min_flow[0]=INF;//源点的第二个分量无穷大    while(1)//标号然后增广，重复有限次可以退出循环    {        for(int i=0; i<maxn_N; i++)            prev[i]=-2;        prev[0]=-1;        q_star=q_end=0;        queue[q_end++]=0;//源点入队        while(q_star<q_end&&prev[t]==-2)        {            int v=queue[q_star++];            for(int i=0; i<=t; i++)            {                if(prev[i]==-2&&(p=customer[v][i]-flow[v][i]))                {                    prev[i]=v;                    queue[q_end++]=i;                    min_flow[i]=min(min_flow[v],p);                }            }        }//        printf("***\n");        if(prev[t]==-2)//没找到去汇点的增广路            break;        for(int i=prev[t],j=t; i!=-1; j=i,i=prev[i])        {            flow[i][j]+=min_flow[t];            flow[j][i]=-flow[i][j];        }    }    int sum=0;    for(int i=0; i<t; i++)        sum+=flow[i][t];    printf("%d\n",sum);}int main(){    init();    ford();    return 0;}