Minimum Cost（最小费用最大流，好题）

Minimum Cost

http://poj.org/problem?id=2516

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 19019		Accepted: 6716

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

Source

POJ Monthly--2005.07.31, Wang Yijie

 

 

 

　　因为每种物品是独立的，所以可以把每种物品拆开来算，再判断最大流是否符合要求即可

 

 #include<iostream>
 #include<algorithm>
 #include<queue>
 #include<cstring>
 using namespace std;

 const int INF=0x3f3f3f3f;
 const int N=;
 const int M=;
 int top;
 int dist[N],pre[N];
 bool vis[N];
 int c[N];
 int maxflow;

 struct Vertex{
     int first;
 }V[N];
 struct Edge{
     int v,next;
     int cap,flow,cost;
 }E[M];

 void init(){
     memset(V,-,sizeof(V));
     top=;
     maxflow=;
 }

 void add_edge(int u,int v,int c,int cost){
     E[top].v=v;
     E[top].cap=c;
     E[top].flow=;
     E[top].cost=cost;
     E[top].next=V[u].first;
     V[u].first=top++;
 }

 void add(int u,int v,int c,int cost){
     add_edge(u,v,c,cost);
     add_edge(v,u,,-cost);
 }

 bool SPFA(int s,int t,int n){
     int i,u,v;
     queue<int>qu;
     memset(vis,false,sizeof(vis));
     memset(c,,sizeof(c));
     memset(pre,-,sizeof(pre));
     for(i=;i<=n;i++){
         dist[i]=INF;
     }
     vis[s]=true;
     c[s]++;
     dist[s]=;
     qu.push(s);
     while(!qu.empty()){
         u=qu.front();
         qu.pop();
         vis[u]=false;
         for(i=V[u].first;~i;i=E[i].next){
             v=E[i].v;
             if(E[i].cap>E[i].flow&&dist[v]>dist[u]+E[i].cost){
                 dist[v]=dist[u]+E[i].cost;
                 pre[v]=i;
                 if(!vis[v]){
                     c[v]++;
                     qu.push(v);
                     vis[v]=true;
                     if(c[v]>n){
                         return false;
                     }
                 }
             }
         }
     }
     if(dist[t]==INF){
         return false;
     }
     return true;
 }

 int MCMF(int s,int t,int n){
     int d;
     int i,mincost;
     mincost=;
     while(SPFA(s,t,n)){
         d=INF;
         for(i=pre[t];~i;i=pre[E[i^].v]){
             d=min(d,E[i].cap-E[i].flow);
         }
         maxflow+=d;
         for(i=pre[t];~i;i=pre[E[i^].v]){
             E[i].flow+=d;
             E[i^].flow-=d;
         }
         mincost+=dist[t]*d;
     }
     return mincost;
 }

 int seller[][];
 int storage[][];
 int matrix[][][];

 int main(){
     int n,m,k;
     int v,u,w,c;
     int s,t;
     while(~scanf("%d %d %d",&n,&m,&k)){
         if(!n&&!m&&!k) break;
         int sum=;
         for(int i=;i<=n;i++){
             for(int j=;j<=k;j++){
                 scanf("%d",&seller[i][j]);
                 sum+=seller[i][j];
             }
         }
         for(int i=;i<=m;i++){
             for(int j=;j<=k;j++){
                 scanf("%d",&storage[i][j]);
             }
         }
         for(int i=;i<=k;i++){
             for(int j=;j<=n;j++){
                 for(int w=;w<=m;w++){
                     scanf("%d",&matrix[i][j][w]);
                 }
             }
         }
         s=,t=n+m+;
         int ANS=;
         int flow=;
         for(int i=;i<=k;i++){
             init();
             for(int j=;j<=n;j++){
                 add(s,j,seller[j][i],);
             }
             for(int j=;j<=n;j++){
                 for(int w=;w<=m;w++){
                     add(j,n+w,storage[w][i],matrix[i][j][w]);
                 }
             }
             for(int j=;j<=m;j++){
                 add(n+j,t,storage[j][i],);///INF
             }
             int ans=MCMF(s,t,t+);
             ANS+=ans;
             flow+=maxflow;
         }

         if(flow==sum) printf("%d\n",ANS);
         else printf("-1\n");
     }
 }