poj 2516

 

Minimum Cost

Time Limit: 4000MS Memory Limit: 65536KTotal Submissions: 8085 Accepted: 2653

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

分析：这题既可以用KM算法，又可以用最大流最小费用。。。两种算法我都不会耶，于是。。。拿来当练习。。。

KM算法：

#include <stdio.h>const int MAX = 51;const int MAXX= 151;int n,m,k,nx,ny,slack;int get[MAX] [MAX],order[MAX][MAX],store[MAX][MAX];int w[MAXX][MAXX],lx [MAXX], ly [MAXX],match [MAXX],can[MAXX];bool sx [MAXX], sy [MAXX],flag;inline int max(int a,int b){return(a>b?a:b);}inline int min(int a,int b){return(a<b?a:b);}inline void makegraph(int x){ int i,j; nx=0,ny=0; for(i=1; i<=n; ++i) for(j=1; j<=order[i][x]; ++j)lx[++nx]=i; for(i=1; i<=m; ++i) for(j=1; j<=store[i][x]; ++j)ly[++ny]=i; for(i=1; i<=nx; ++i) for(j=1; j<=ny; ++j)w[i][j]=get[lx[i]][ly[j]];}bool path(int u){ sx[u]=1; for(int v = 1;v <=ny;++v) { if(sy[v]) continue; int temp = w[u][v] - lx[u] - ly[v]; if(temp == 0) { sy[v] = 1; if(!match[v] || path(match[v])) { can[u]=v; match[v] = u; return 1; } } else slack =min(slack,temp); } return 0;}int KM(){ int i,j; for(i=1;i<=nx;++i) { lx[i]=0x7FFFFFFF; for(j=1;j<=ny;++j)match[j]=0,ly[j]=0,lx[i]=min(lx[i],w[i][j]); } for(int u=1;u <=nx;++u) while(1) { for(i=1;i<=nx;++i)sx[i]=0; for(i=1;i<=ny;++i)sy[i]=0; slack=0x7FFFFFFF; if(path(u)) break; for(i=1;i<=nx;++i) if(sx[i])lx[i]+=slack; for(j=1;j<=ny;++j) if(sy[j])ly[j]-=slack; } int sum=0; for(i=1; i<=nx; ++i)sum+=w[i][can[i]]; return sum;}inline int in(){ char ch; int a=0; while(!(((ch=getchar())>='0')&&(ch<='9'))); a*=10;a+=ch-'0'; while(((ch=getchar())>='0')&&(ch<='9'))a*=10,a+=ch-'0'; return a;}int main(){ int i,j,h,cost,s1,s2; while(n=in(),m=in(),k=in()) { for(i=1;i<=k;++i)can[i]=0; for(i=1; i<=n; ++i) for(j=1; j<=k; ++j)order[i][j]=in(); for(i=1; i<=m; ++i) for(j=1; j<=k; ++j)store[i][j]=in(); flag=1; for(j=1;j<=k;++j) { s1=s2=0; for(i=1;i<=n;++i)s1+=order[i][j]; for(i=1;i<=m;++i)s2+=store[i][j]; if(s1>s2) { flag=0; break; } } cost=0; for(h=1; h<=k; ++h) { for(i=1; i<=n; ++i) for(j=1; j<=m; ++j)get[i][j]=in(); if(flag)makegraph(h),cost+=KM(); } printf("%d/n",(flag?cost:-1)); } return 0;} 

最小费用流：

#include <stdio.h>#include <string.h>const int MN=52;const int MM=110;const int INF=0x7f7f7f7f;int n,m,k;int order[MN][MN],store[MN][MN],mat[MN][MN];int queue[MM],flow[MM][MM],c[MM][MM],cost[MM][MM],pre[MM],dis[MM],mf[MM];bool visit[MM],flag;inline int MIN(int x,int y){ return x<y?x:y;}inline void get(int& a){ char ch=getchar(); while (ch<'0'||ch>'9') ch=getchar(); for (a=0; ch>='0'&&ch<='9'; ch=getchar()) a=a*10+ch-48;}inline void ConstructMap(int p,int t){ int i,j,k; memset(c,0,sizeof(c)); memset(cost,0,sizeof(cost)); for(i=1; i<=m; ++i)c[0][i]=store[i][p]; for(i=1; i<=m; ++i) for(j=1; j<=n; ++j) { k=j+m; c[i][k]=INF; cost[i][k]=mat[j][i]; cost[k][i]=-mat[j][i]; } for (i=1; i<=n; ++i)c[i+m][t]=order[i][p];}inline bool spfa( int n, int s, int t ){ int top,tail,u,i,v; memset(dis,0x7f,sizeof(dis)); memset(visit,1,sizeof(visit)); memset(pre,-1,sizeof(pre)); dis[s]=0,top=0,tail=1,queue[0]=s,visit[s]=0,mf[s]=INF; while(top!=tail) { u=queue[top],visit[u]=1; ++top; if (top==n)top=0; for(v=0; v<n;++v) if(c[u][v]>flow[u][v]&&dis[v]-cost[u][v]>dis[u]) { dis[v]=dis[u]+cost[u][v]; mf[v]=MIN(mf[u],c[u][v]-flow[u][v]); pre[v]=u; if(visit[v]) { queue[tail]=v,visit[v]=0; ++tail; if(tail==n)tail=0; } } } return(pre[t]>=0);}inline int Min_Cost_Max_Flow(int n,int s,int t){ int i, j,u,v,ans=0; memset(flow,0,sizeof(flow)); while(spfa(n,s,t)) { for(u=pre[v=t];v!=0;v=u,u=pre[v]) { flow[u][v]+=mf[t]; flow[v][u]-=mf[t]; } ans+=mf[t]*dis[t]; } return ans;}int main(){ int i,j,p,sum,s1,s2; while(get(n),get(m),get(k),k) { for(i=1;i<=n;++i) for(j=1;j<=k;++j) get(order[i][j]); for(i=1;i<=m;++i) for(j=1;j<=k;++j) get(store[i][j]); sum=0,flag=1; for(j=1;j<=k;++j) { s1=s2=0; for(i=1;i<=n;++i)s1+=order[i][j]; for(i=1;i<=m;++i)s2+=store[i][j]; if(s1>s2) { flag=0; break; } } for(p=1;p<=k;++p) { for(i=1;i<=n;++i) for(j=1;j<=m;++j) get(mat[i][j]); if (flag)ConstructMap(p,n+m+1),sum+=Min_Cost_Max_Flow(n+m+2,0,n+m+1); } printf("%d/n",(flag?sum:-1)); } return 0;}