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BNUOJ 1268 PIGS

时间:2014-08-07 22:05:08      阅读:322      评论:0      收藏:0      [点我收藏+]

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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

 
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。
 
 
 
 
bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <vector>
 6 #include <climits>
 7 #include <algorithm>
 8 #include <cmath>
 9 #include <queue>
10 #define LL long long
11 #define INF 0x3f3f3f3f
12 using namespace std;
13 const int maxn = 1005;
14 int n,m,a[maxn],p[maxn];
15 int arc[maxn][maxn],pre[maxn],pigs[maxn];
16 int bfs(){
17     int u,v,f = 0;
18     queue<int>q;
19     while(true){
20         while(!q.empty()) q.pop();
21         memset(a,0,sizeof(a));
22         memset(p,0,sizeof(p));
23         a[0] = INF;
24         q.push(0);
25         while(!q.empty()){
26             u = q.front();
27             q.pop();
28             for(v = 0; v <= n+1; v++){
29                 if(!a[v] && arc[u][v] > 0){
30                     a[v] = min(a[u],arc[u][v]);
31                     p[v] = u;
32                     q.push(v);
33                 }
34             }
35             if(a[n+1]) break;
36         }
37         if(!a[n+1]) break;
38         for(u = n+1; u; u = p[u]){
39             arc[p[u]][u] -= a[n+1];
40             arc[u][p[u]] += a[n+1];
41         }
42         f += a[n+1];
43     }
44     return f;
45 }
46 int main(){
47     int i,j,k,buy,e;
48     while(~scanf("%d%d",&m,&n)){
49         memset(arc,0,sizeof(arc));
50         memset(pre,0,sizeof(pre));
51         for(i = 1; i <= m; i++)
52             scanf("%d",pigs+i);
53         for(i = 1; i <= n; i++){
54             scanf("%d",&k);
55             while(k--){
56                 scanf("%d",&e);
57                 if(pre[e]) arc[pre[e]][i] = INF;
58                 else arc[pre[e]][i] += pigs[e];
59                 pre[e] = i;
60             }
61             scanf("%d",&buy);
62             arc[i][n+1] += buy;
63         }
64         cout<<bfs()<<endl;
65     }
66     return 0;
67 }
View Code

 

BNUOJ 1268 PIGS,布布扣,bubuko.com

BNUOJ 1268 PIGS

标签:style   blog   http   color   java   os   io   for   

原文地址:http://www.cnblogs.com/crackpotisback/p/3897985.html

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