POJ1149_PIGS(网络流/EK)

Posted 2021-11-28 ldxsuanfa

PIGS

Time Limit:?1000MS	?	Memory Limit:?10000K
Total Submissions:?15721	?	Accepted:?7021

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

解题报告

昨天開始学网络流，这是第一题网络流建图的题。

题目意思：

养猪场M个猪圈。每一个猪圈都上锁，主人又没有钥匙。N个顾客买猪。且每一个顾客有一些猪圈的钥匙（这是什么情况，主人没有钥匙。反而买主有钥匙。sad...）

一天，要到养猪场买猪的顾客都会提前告诉养猪场主人。包含拥有的钥匙。买几头猪。养猪场主人能够安排销售计划使得卖出去的猪数目最大。

每当顾客来了。会把他拥有钥匙的猪圈全都打开，养猪场主人挑一些猪买出去，养猪场主人还能够又一次分配被打开猪圈的猪。

猪圈能够容纳猪的数量不限。

思路：

由于一開始猪圈是上锁的。所以把顾客其中转站，另设两节点，源点和汇点。

源点和每一个猪圈的第一个顾客连边。边的权值是猪圈里的猪的数目。

顾客j紧跟着顾客i打开某猪圈，则<i,j>的权值是+oo，表示假设顾客j在顾客i之后打开猪圈，主人能够跟据顾客j的需求把其它猪圈的猪赶到该猪圈，这样顾客j就能够买到尽可能多的猪。

每一个顾客和汇点相连。边权是每一个顾客的需求量。

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#define inf 99999999
#define M 10100
#define N 1100
using namespace std;
int pigh[M],edge[N][N],p[N],a[N],pre[N],m,n,flow;
queue<int >Q;
void ek()
{
    while(1)
    {
        while(!Q.empty())Q.pop();
        memset(a,0,sizeof(a));
        memset(p,0,sizeof(p));
        a[0]=inf;
        Q.push(0);
        while(!Q.empty())
        {
            int u=Q.front();
            Q.pop();
            for(int v=0;v<=n+1;v++)
            {
                if(!a[v]&&edge[u][v]>0)
                {
                    a[v]=min(a[u],edge[u][v]);
                    p[v]=u;
                    Q.push(v);
                }
            }
            if(a[n+1])break;
        }
        if(!a[n+1])break;
        for(int u=n+1;u!=0;u=p[u])
        {
            edge[p[u]][u]-=a[n+1];
            edge[u][p[u]]+=a[n+1];
        }
        flow+=a[n+1];
    }
}
int main()
{
    int i,j,k,u,b;
    while(~scanf("%d%d",&m,&n))
    {
        flow=0;
        memset(pigh,0,sizeof(pigh));
        memset(edge,0,sizeof(edge));
        memset(pre,0,sizeof(pre));
        for(i=1; i<=m; i++)
            scanf("%d",&pigh[i]);
        for(i=1; i<=n; i++)
        {
            scanf("%d",&k);
            while(k--)
            {
                scanf("%d",&u);
                if(!pre[u])
                {
                    edge[pre[u]][i]+=pigh[u];
                    pre[u]=i;
                }
                else
                {
                    edge[pre[u]][i]=inf;
                    pre[u]=i;
                }
            }
            scanf("%d",&b);
            edge[i][n+1]+=b;
        }
        ek();
//        for(i=0;i<=n+1;i++)
//        {
//            for(j=0;j<=n+1;j++)
//            {
//                cout<<edge[i][j]<<" ";
//            }
//            cout<<endl;
//        }
        printf("%d\n",flow);
    }
    return 0;
}