POJ1149 PIGS （网络流）-白红宇

POJ1149 PIGS （网络流）

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发布时间：2019-06-21

本文共 2792 字，大约阅读时间需要 9 分钟。

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20146		Accepted: 9218

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 33 1 102 1 2 22 1 3 31 2 6

Sample Output

7 【分析】直接套的书上的标号法模板。

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                   #define inf 0x3f3f3f3f#define mod 1000000007typedef long long ll;using namespace std;const int N=105;const int M=1005;int s,t,n,m,num,k;int customer[N][N];int flow[N][N];int house[M],last[M];int pre[N],minflow[N];void Build() { memset(last,0,sizeof(last)); memset(customer,0,sizeof(customer)); scanf("%d%d",&m,&n); s=0; t=n+1; for(int i=1; i<=m; i++)scanf("%d",&house[i]); for(int i=1; i<=n; i++) { scanf("%d",&num); for(int j=0; j

q; int p=0; memset(flow,0,sizeof(flow)); minflow[0]=inf; while(1) { while(!q.empty())q.pop(); for(int i=0; i

=0; j=i,i=pre[i]) { flow[i][j]+=minflow[t]; flow[j][i]=-flow[i][j]; } } for(int i=0;i

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转载于:https://www.cnblogs.com/jianrenfang/p/5826314.html

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