PIGS 分类： POJ 图论 2015-08-10 09:15 3人阅读 评论(0) 收藏

PIGS
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 18209 Accepted: 8277

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

Source
Croatia OI 2002 Final Exam - First day
题意:有不同的猪圈,买家手里有着猪圈的钥匙,拥有那个猪圈的钥匙就可以买这个猪圈的猪,怎样安排买猪才能卖出更多的猪;

#include <map>
#include <list>
#include <climits>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#define WW freopen("output.txt","w",stdout)

const int Max = 100000;

int cus[110][110];//图的容量

int flow[110][110];//图的流量

int num[1100];//猪的数量

int pre[1100];//前一个买家

int n,m,data;

int t,s;

int ford()//一般增广路算法
{
    int minflow[110];
    queue<int >Q;
    memset(flow,0,sizeof(flow));
    minflow[0] = INF;
    t=n+1;
    s=0;
    while(1)
    {
        for(int i=0;i<=t;i++)
        {
            pre[i]=-2;
        }
        while(!Q.empty())
        {
            Q.pop();
        }
        pre[0]=-1;
        Q.push(0);
        while(!Q.empty()&&pre[t]==-2)//BFS找增广路
        {
            int v=Q.front();
            int p;
            Q.pop();
            for(int i=0;i<=t;i++)
            {
                if(pre[i]==-2&&(p=cus[v][i]-flow[v][i]))
                {
                    pre[i]=v;
                    Q.push(i);
                    minflow[i]=min(minflow[v],p);
                }
            }
        }
        if(pre[t]==-2)//如果找不到汇点,则说明图中不存在增广路,算法结束
        {
            break;
        }
        for(int i=pre[t],j=t;i!=-1;j=i,i=pre[i])//回溯进行增广
        {
            flow[i][j] +=minflow[t];
            flow[j][i]-=minflow[t];
        }
    }
    int sum=0;
    for(int i=0;i<t;i++)
    {
        sum+=flow[i][t];//每个点到汇点的总和为最大流
    }
    return sum;
}

int main()
{
    while(~scanf("%d %d",&m,&n))
    {
        memset(pre,-1,sizeof(pre));
        memset(cus,0,sizeof(cus));
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&num[i]);
        }
        int s;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&data);
            for(int j=0;j<data;j++)
            {
                scanf("%d",&s);
                if(pre[s]==-1)//如果这个猪圈之前没有人,则这个人是第一个人,所以与源点之间的容量为猪圈的猪的数量
                {
                    cus[0][i]+=num[s];
                }
                else
                {
                    cus[pre[s]][i]=INF;//如果已经有人买了,所以他必须在这个人之后买,但两人之间又没有关系,所以容量为INF(无穷大)
                }
                pre[s]=i;
            }
            scanf("%d",&cus[i][n+1]);//到汇点的容量为每个人想买猪的数量
        }
        printf("%d\n",ford());
    }

    return 0;
}

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