PIGS
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 18209 Accepted: 8277
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
题意:有不同的猪圈,买家手里有着猪圈的钥匙,拥有那个猪圈的钥匙就可以买这个猪圈的猪,怎样安排买猪才能卖出更多的猪;
#include <map>
#include <list>
#include <climits>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#define WW freopen("output.txt","w",stdout)
const int Max = 100000;
int cus[110][110];//图的容量
int flow[110][110];//图的流量
int num[1100];//猪的数量
int pre[1100];//前一个买家
int n,m,data;
int t,s;
int ford()//一般增广路算法
{
int minflow[110];
queue<int >Q;
memset(flow,0,sizeof(flow));
minflow[0] = INF;
t=n+1;
s=0;
while(1)
{
for(int i=0;i<=t;i++)
{
pre[i]=-2;
}
while(!Q.empty())
{
Q.pop();
}
pre[0]=-1;
Q.push(0);
while(!Q.empty()&&pre[t]==-2)//BFS找增广路
{
int v=Q.front();
int p;
Q.pop();
for(int i=0;i<=t;i++)
{
if(pre[i]==-2&&(p=cus[v][i]-flow[v][i]))
{
pre[i]=v;
Q.push(i);
minflow[i]=min(minflow[v],p);
}
}
}
if(pre[t]==-2)//如果找不到汇点,则说明图中不存在增广路,算法结束
{
break;
}
for(int i=pre[t],j=t;i!=-1;j=i,i=pre[i])//回溯进行增广
{
flow[i][j] +=minflow[t];
flow[j][i]-=minflow[t];
}
}
int sum=0;
for(int i=0;i<t;i++)
{
sum+=flow[i][t];//每个点到汇点的总和为最大流
}
return sum;
}
int main()
{
while(~scanf("%d %d",&m,&n))
{
memset(pre,-1,sizeof(pre));
memset(cus,0,sizeof(cus));
for(int i=1;i<=m;i++)
{
scanf("%d",&num[i]);
}
int s;
for(int i=1;i<=n;i++)
{
scanf("%d",&data);
for(int j=0;j<data;j++)
{
scanf("%d",&s);
if(pre[s]==-1)//如果这个猪圈之前没有人,则这个人是第一个人,所以与源点之间的容量为猪圈的猪的数量
{
cus[0][i]+=num[s];
}
else
{
cus[pre[s]][i]=INF;//如果已经有人买了,所以他必须在这个人之后买,但两人之间又没有关系,所以容量为INF(无穷大)
}
pre[s]=i;
}
scanf("%d",&cus[i][n+1]);//到汇点的容量为每个人想买猪的数量
}
printf("%d\n",ford());
}
return 0;
}
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