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
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<deque>
using namespace std;
#define maxn 550
#define INF 99999999
int Map[maxn][maxn];
int prev[maxn];
bool vis[maxn];
int come[1100],pig[1100];
void EK(int x,int n)
{
int sum=0;
int temp,u,v,i;
while(1)
{
memset(vis,false,sizeof(vis));
memset(prev,0,sizeof(prev));
deque<int>Q;
prev[x]=0;
vis[x]=true;
Q.push_back(x);
while(!Q.empty())
{
// printf("**********\n");
u=Q.front();
Q.pop_front();
for(v=1;v<=n;v++)
{
//printf("+++++++++++\n");
if(!vis[v]&&Map[u][v])
{
prev[v]=u;
vis[v]=true;
if(v==n)
{
Q.clear();
break;
}
else
Q.push_back(v);
}
}
}
if(!vis[n])
break;
temp=INF;
i=n;
while(prev[i])
{
if(temp>Map[prev[i]][i])
temp=Map[prev[i]][i];
//printf("+++++++++++\n");
i=prev[i];
}
i=n;
while(prev[i])
{
Map[prev[i]][i]-=temp;
Map[i][prev[i]]+=temp;
// printf("**********\n");
i=prev[i];
}
//printf("%d\n",temp);
sum+=temp;
}
printf("%d\n",sum);
}
int main()
{
int m,n;
int i,j,k;
int t;
int x;
while(scanf("%d%d",&m,&n)!=EOF)
{
k=n+1;
memset(come,0,sizeof(come));
memset(Map,0,sizeof(Map));
for(i=1;i<=m;i++)
scanf("%d",&pig[i]);
for(i=1;i<=n;i++)
{
scanf("%d",&t);
for(j=1;j<=t;j++)
{
scanf("%d",&x);
if(!come[x])
Map[k][i]+=pig[x];
else
Map[come[x]][i]=INF;
come[x]=i;//记录猪圈的指向
}
scanf("%d",&Map[i][k+1]);
}
EK(k,k+1);
}
return 0;
}
