poj 2516
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 8085 | Accepted: 2653 |
Description
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
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
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; }