poj2516 最小费用流

 

如题：http://poj.org/problem?id=2516

 

Minimum Cost

Time Limit: 4000MS Memory Limit: 65536KTotal Submissions: 14338 Accepted: 4909

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 10 1 11 2 21 0 11 2 31 1 12 1 11 1 132200 0 0

Sample Output

4-1

Source

POJ Monthly--2005.07.31, Wang Yijie

 

 

 

题意：有M个供应点，N个零售商，K种商品，给出N个零售商K种商品分别需要的数量和M个供应点K种商品分别的数量，然后给出K个矩阵，第i个矩阵第j行k列表示每一单位的第i种商品从k提供商到j零售商需要的花费。要求输出最小总花费。

 

思路：最小费用流建立K个图处理K次，（1《i《k）每次从0到M个供应商建立权值为拥有第i种产品的数量，从M个供应商到M+j表示从供应商到零售商j建立一条拥有i产品的数量

和花费的边，然后零售商到汇点（N+M+1）建立需要i产品数量的边。用队列优化的bellman_ford每次求出一条最短，然后计算流，一共K次。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
#define MAX 200
#define INF 0x0fffffff
#define min(a,b)(a<b?a:b)

int N,M,K;
int need[MAX][MAX]; //need[i][j] i客户需要j产品的数量
int hav[MAX][MAX]; //hav[i][j]  提供点i有j产品的数量
int money[MAX][MAX][MAX]; //money[i][j][k]  将i产品从k运送到j客户需要的钱
int sum_hav[MAX];
int sum_need[MAX];
int cap[MAX][MAX];
int cost[MAX][MAX];
int flow[MAX][MAX];
int pre[MAX];
int dist[MAX];

void spfa(int s,int t)
{
 int i;
 queue<int>que;
 for(i=0;i<=t;i++)
  dist[i]=INF;
 int inq[MAX]={0};
 memset(pre,-1,sizeof(pre));
 dist[s]=0;
 que.push(s);
 inq[s]=1;
 while(!que.empty())
 {
  int u=que.front();
  que.pop();
  inq[u]=0;
  for(i=0;i<=t;i++)
   if(cap[u][i]>flow[u][i]&&dist[i]>dist[u]+cost[u][i])
   {
    dist[i]=dist[u]+cost[u][i];
    pre[i]=u;
    if(!inq[i])
    {
     inq[i]=1;
     que.push(i);
    }
   }
 }
}
int min_flow_cost(int s,int t)
{
 memset(flow,0,sizeof(flow));
 int v;
 int res=0;
 while(1)
 {
  spfa(s,t);
  if(pre[t]==-1)
   break;
  int d=INF;
  for(v=t;v!=s;v=pre[v])
  {
   d=min(d,cap[pre[v]][v]-flow[pre[v]][v]);
  }
  res+=d*dist[t];
  for(v=t;v!=s;v=pre[v])
  {
   flow[pre[v]][v]+=d;
   flow[v][pre[v]]-=d;
  }
 }
 return res;
}
int main()
{
// freopen("C:\\1.txt","r",stdin);
  while(~scanf("%d%d%d",&N,&M,&K))
 {
  if(!N&&!M&&!K)
   break;
  int i,j,k;
  int st=0,ed=N+M+1;
  memset(sum_hav,0,sizeof(sum_hav));
  memset(sum_need,0,sizeof(sum_need));
  for(i=1;i<=N;i++)
   for(j=1;j<=K;j++)
   {
    scanf("%d",&need[i][j]);
    sum_need[j]+=need[i][j];
   }
  for(i=1;i<=M;i++)
   for(j=1;j<=K;j++)
   {
    scanf("%d",&hav[i][j]);
    sum_hav[j]+=hav[i][j];
   }
  for(i=1;i<=K;i++)
   for(j=1;j<=N;j++)
    for(k=1;k<=M;k++)
     scanf("%d",&money[i][j][k]);
   int flag=0;
  for(i=1;i<=K;i++)
   if(sum_hav[i]<sum_need[i])
   {
    printf("-1\n");
    flag=1;
    break;
   }
  if(flag)
   continue;
  int flow=0;
  for(i=1;i<=K;i++)
  {
   memset(cap,0,sizeof(cap));
   memset(cost,0,sizeof(cost));
   for(j=1;j<=M;j++)
    for(k=1;k<=N;k++)
    {
     cap[j][k+M]=hav[j][i];
     cost[j][k+M]=money[i][k][j];
     cost[k+M][j]=-cost[j][k+M];
    }
   for(j=1;j<=M;j++)
    cap[st][j]=hav[j][i];
   for(j=1;j<=N;j++)
    cap[j+M][ed]=need[j][i];
   flow+=min_flow_cost(st,ed);
  }
  printf("%d\n",flow);
 }
 return 0;
}