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poj 2516 Minimum Cost

2024-07-13 16:01:37   223人阅读

Minimum Cost
Time Limit: 4000MS Memory Limit: 65536K
Total Submissions: 13242 Accepted: 4500

Description

Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport. 

It‘s known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places‘ storage of K kinds of goods, N shopkeepers‘ order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, K (0 < N, M, K < 50), which are described above. The next N lines give the shopkeepers‘ orders, with each line containing K integers (there integers are belong to [0, 3]), which represents the amount of goods each shopkeeper needs. The next M lines give the supply places‘ storage, with each line containing K integers (there integers are also belong to [0, 3]), which represents the amount of goods stored in that supply place. 

Then come K integer matrices (each with the size N * M), the integer (this integer is belong to (0, 100)) at the i-th row, j-th column in the k-th matrix represents the cost to transport one unit of k-th goods from the j-th supply place to the i-th shopkeeper. 

The input is terminated with three "0"s. This test case should not be processed.

Output

For each test case, if Dearboy can satisfy all the needs of all the shopkeepers, print in one line an integer, which is the minimum cost; otherwise just output "-1".

Sample Input

1 3 3   
1 1 1
0 1 1
1 2 2
1 0 1
1 2 3
1 1 1
2 1 1

1 1 1
3
2
20

0 0 0

Sample Output

4
-1


题意：n个客户，m个仓库，l件物品。

n行l列 表示n个客户对l件物品的需求量

m行l列 表示m个仓库对l件物品的的供给量

l行 表示l个物品从i仓库到j客户的一个物品的运费价格

判断是否可以满足客户需求，如果满足求出最小的运费。

思路：最小费用最大流问题

将l件物品分开求最小费用然后相加。
对每个仓库是一个结点，每个客户也是一个结点，除此之外再加上s源点和t结束点

代码（附解释）如下：

#include<iostream>  
#include<cstdio>  
#include<algorithm> 
#include<queue>  
using namespace std;  
#define min(a,b) a<b?a:b  
#define N 130  
#define INF 999999  
int need[N][N],offer[N][N],sum_need[N],sum_offer[N];   
int map[N][N],cost[N][N];  
int n,m,l;  
int pre[N];  
int s,t;  //源点和汇点  

int spfa()  
{  
    int dis[N],vis[N];   
    int i;  
    queue<int>q;  
    for(i=0; i<=t; i++)   
        dis[i]=INF;
	
	memset(vis,0,sizeof(vis));
    
	vis[s]=1;  
    dis[s]=0;  
    q.push(s);  
    while(!q.empty())  
    { 
        int k=q.front();  
		q.pop();  
        vis[k]=0;  
        for(i=0; i<=t; i++)  
        {  
            if(map[k][i] && dis[i]>dis[k]+cost[k][i])  
            {  
                dis[i]=dis[k]+cost[k][i];  
                pre[i]=k;  
                if(vis[i]==0)  
                {  
                    vis[i]=1;  
                    q.push(i);  
                }  
            }  
        } 
	}
	if(dis[t]!=INF)  
		return 1;  
	else  
		return 0;  
}  

int fond()  
{  
    int i,j;  
    int Min=INF;  
    j=0;  
    while(spfa())  
    {  
        for(i=t;i!=s; i=pre[i])  
            Min=min(Min,map[pre[i]][i]);  
        for(i=t; i!=s; i=pre[i])  
        {  
            map[pre[i]][i]-=Min;  
            map[i][pre[i]]+=Min;  
			j+=cost[pre[i]][i]*Min;
        }  
    }  
    return j;  
} 

int main()  
{  
    int i,j,k,sum,sign;  
    while(scanf("%d%d%d",&n,&m,&l))  // n个客户，m个仓库，l件物品  
    {  
        if(n==0 && m==0 && l==0)  break;  
        sum=0;  
        memset(sum_need,0,sizeof(sum_need));  
        memset(sum_offer,0,sizeof(sum_offer));  
        
		for(i=1; i<=n; i++)    //求客户对物品总的需求量
            for(j=1; j<=l; j++)  
            {  
                scanf("%d",&need[i][j]);  
                sum_need[j]+=need[i][j];  
            } 
			
			for(i=1;i<=m;i++)    //求仓库对物品总的供给量
				for(j=1;j<=l;j++)  
				{  
					scanf("%d",&offer[i][j]);  
					sum_offer[j]+=offer[i][j];  
				}
				
				sign=0;  
				for(i=1;i<=l;i++)  
					if(sum_offer[i]<sum_need[i])   //供给小于需求
					{  
						sign=1;  
						break;  
					}   
					
					s=0;       // 源点
					t=m+n+1;   // 汇点(把客户和仓库都看成节点) 
					
					for(k=1;k<=l;k++)   // 把l件物品分开计算
					{  
						for(i=0;i<=t; i++)  
							for(j=0;j<=t;j++)  
								map[i][j]=cost[i][j]=0;  //初始化 节点是0,1,2....,n
						
						for(i=1+m;i<=n+m;i++)
							for(j=1;j<=m;j++)  // 注意  把客户节点是m+1,m+2,....,m+n
								   
								{  
									scanf("%d",&cost[j][i]);   
									cost[i][j]=-cost[j][i];  //注意反向
								} 
								if(sign==1)  continue;    
								
								for(i=1;i<=m;i++)        
									map[s][i]=offer[i][k]; // 源点s到仓库的边权值为仓库的供给量
								
								for(i=1;i<=m; i++)     
									for(j=1+m; j<=n+m; j++)  
										map[i][j]=offer[i][k];  //仓库到客户的的边权值为客户的需求量
									
									for(i=m+1; i<=m+n; i++)      
										map[i][t]=need[i-m][k];  //客户到汇点t的边权值为客户的需求量
									
									sum+=fond();  // 对k件物品从仓库到客户的最小费用求和
					} 
					
					if(sign==1)  
						printf("-1\n");  
					else  
						printf("%d\n",sum);  
    }   
    return 0;  
}





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