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

Sample Output

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 = ;

 int n,m,a[maxn],p[maxn];

 int arc[maxn][maxn],pre[maxn],pigs[maxn];

 int bfs(){

     int u,v,f = ;

     queue<int>q;

     while(true){

         while(!q.empty()) q.pop();

         memset(a,,sizeof(a));

         memset(p,,sizeof(p));

         a[] = INF;

         q.push();

         while(!q.empty()){

             u = q.front();

             q.pop();

             for(v = ; v <= n+; v++){

                 if(!a[v] && arc[u][v] > ){

                     a[v] = min(a[u],arc[u][v]);

                     p[v] = u;

                     q.push(v);

                 }

             }

             if(a[n+]) break;

         }

         if(!a[n+]) break;

         for(u = n+; u; u = p[u]){

             arc[p[u]][u] -= a[n+];

             arc[u][p[u]] += a[n+];

         }

         f += a[n+];

     }

     return f;

 }

 int main(){

     int i,j,k,buy,e;

     while(~scanf("%d%d",&m,&n)){

         memset(arc,,sizeof(arc));

         memset(pre,,sizeof(pre));

         for(i = ; i <= m; i++)

             scanf("%d",pigs+i);

         for(i = ; 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+] += buy;

         }

         cout<<bfs()<<endl;

     }

     return ;

 }

巴特西