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
#include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; #define MAXN 1010 #define INF 9999999 int map[MAXN][MAXN]; int f[MAXN]; int pig[MAXN]; int book[MAXN]; int a[MAXN]; int EK(int n) { int i,sum=0,flog; while(1) { memset(a,0,sizeof(a)); memset(book,0,sizeof(book)); queue<int> q; f[0]=INF; q.push(0); flog=0; while(q.size()&&!flog) { int k=q.front(); q.pop(); for(i=1; i<=n; i++) { if(book[i]==0&&map[k][i]>0) { book[i]=1; a[i]=k; f[i]=min(map[k][i],f[k]); q.push(i); if(i==n) { flog=1; break; } } } } if(flog==0) break; for(i=n; i!=0; i=a[i]) { int l=a[i]; map[l][i]-=f[n]; map[i][l]+=f[n]; } sum=sum+f[n]; } printf("%d\n",sum); } int main() { int m,n,i; int p[MAXN]; while(~scanf("%d %d",&m,&n)) { memset(p,0,sizeof(p)); memset(map,0,sizeof(map)); int a,b,l,k; for(i=1; i<=m; i++) scanf("%d",&pig[i]); for(i=1; i<=n; i++)// 建 图 { scanf("%d",&l); while(l--) { scanf("%d",&k); if(p[k]==0) { map[0][i]+=pig[k]; p[k]=i; } else { map[p[k]][i]=INF; p[k]=i; } } scanf("%d",&k); map[i][n+1]=k; } EK(n+1); } }