BNUOJ 1268 PIGS

PIGS

Time Limit: 1000ms

Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1149
64-bit integer IO format: %lld      Java class name: Main

 

 

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 arives, 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：最大流问题。买猪。建图是关键啊！0为源n+1为汇。把买者看成流网络上的点，最先选择某个猪舍跟源点建立边，容量为这个猪舍猪头数，第二个选择某个猪舍的与第一个选择这个猪舍的买者建立边，容量为无穷大。类推。。。最后把最后与某个猪舍建立边的买者，再与汇建立边，数目为此买者的购买量！为什么这样建图呢？首先是源点！源点到第一个买者，我们考虑的不仅仅是第一个买者的购买力，还要考虑其他人的购买力，所以干脆把整个猪舍的猪都卖掉！假想可以卖完！至于中间的边，就是让尽可能的猪留给后面的人可以来买，但是最后的人的购买力有限，最多只能购买那么多！所以与汇邻接的边的权值为买者的购买量！最后流到汇的就是卖出的猪的头数。

 

之所以可以这样建图，是因为题意是指，第一个人打开某个猪舍后，这个猪舍就不会关了，这里面的猪可以随意卖给其他人。

 

 

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#include <queue>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1005;
int n,m,a[maxn],p[maxn];
int arc[maxn][maxn],pre[maxn],pigs[maxn];
int bfs(){
    int u,v,f = 0;
    queue<int>q;
    while(true){
        while(!q.empty()) q.pop();
        memset(a,0,sizeof(a));
        memset(p,0,sizeof(p));
        a[0] = INF;
        q.push(0);
        while(!q.empty()){
            u = q.front();
            q.pop();
            for(v = 0; v <= n+1; v++){
                if(!a[v] && arc[u][v] > 0){
                    a[v] = min(a[u],arc[u][v]);
                    p[v] = u;
                    q.push(v);
                }
            }
            if(a[n+1]) break;
        }
        if(!a[n+1]) break;
        for(u = n+1; u; u = p[u]){
            arc[p[u]][u] -= a[n+1];
            arc[u][p[u]] += a[n+1];
        }
        f += a[n+1];
    }
    return f;
}
int main(){
    int i,j,k,buy,e;
    while(~scanf("%d%d",&m,&n)){
        memset(arc,0,sizeof(arc));
        memset(pre,0,sizeof(pre));
        for(i = 1; i <= m; i++)
            scanf("%d",pigs+i);
        for(i = 1; i <= n; i++){
            scanf("%d",&k);
            while(k--){
                scanf("%d",&e);
                if(pre[e]) arc[pre[e]][i] = INF;
                else arc[pre[e]][i] += pigs[e];
                pre[e] = i;
            }
            scanf("%d",&buy);
            arc[i][n+1] += buy;
        }
        cout<<bfs()<<endl;
    }
    return 0;
}