http://poj.org/problem?id=2400
KM算法http://philoscience.iteye.com/blog/1754498
题意:每个雇主对雇员有个满意度 雇员对雇主有个满意度 求最匹配的雇员与雇主
最小权匹配 这个给出的数据矩阵貌似是反着的 建图的时候反一下就好了
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<queue> 6 #include<stdlib.h> 7 #define N 110 8 #define INF 0x3f3f3f 9 using namespace std; 10 int w[N][N],n,x[N],y[N],lx[N],ly[N],link[N],sum,o[N],vis[N],num; 11 bool match(int u) 12 { 13 x[u] = 1; 14 for(int i = 1; i <= n ; i++) 15 if(lx[u]+ly[i]==w[u][i]&&!y[i]) 16 { 17 y[i] = 1; 18 if(!link[i]||match(link[i])) 19 { 20 link[i] = u; 21 return true; 22 } 23 } 24 return false; 25 } 26 double km() 27 { 28 int i,j,g; 29 for(i = 1; i <= n ; i++) 30 link[i] = lx[i] = ly[i]=0; 31 for(i = 1 ; i <= n ; i++) 32 { 33 for(;;) 34 { 35 for(j = 1 ;j <= n ; j++) 36 x[j] = y[j] = 0; 37 if(match(i)) break; 38 int d = INF; 39 for(j = 1; j <= n ; j++) 40 { 41 if(x[j]) 42 for(g = 1; g <= n ; g++) 43 if(!y[g]) 44 d = min(d,lx[j]+ly[g]-w[j][g]); 45 } 46 for(j = 1; j <= n ; j++) 47 { 48 if(x[j]) lx[j]-=d; 49 if(y[j]) ly[j]+=d; 50 } 51 } 52 } 53 sum=0; 54 for(i = 1; i <= n ; i++) 55 sum-=(lx[i]+ly[i]); 56 return 1.0*sum/(2*n); 57 } 58 void dfs(int u,int ss) 59 { 60 int i; 61 if(ss>sum) return ; 62 if(u==n+1) 63 { 64 printf("Best Pairing %d ",++num); 65 for(i = 1 ;i <= n ; i++) 66 printf("Supervisor %d with Employee %d ",i,o[i]); 67 return ; 68 } 69 for(i = 1; i <= n ; i++) 70 { 71 if(!vis[i]) 72 { 73 vis[i] = 1; 74 o[u] = i; 75 dfs(u+1,ss-w[u][i]); 76 vis[i] = 0; 77 } 78 } 79 } 80 int main() 81 { 82 int i,j,kk=0,cas,a; 83 scanf("%d",&cas); 84 while(cas--) 85 { 86 kk++;num=0; 87 scanf("%d",&n); 88 memset(w,0,sizeof(w)); 89 memset(vis,0,sizeof(vis)); 90 for(i = 1 ; i <= n ; i++) 91 for(j = 1; j <= n ; j++) 92 { 93 scanf("%d",&a); 94 w[a][i] -= j; 95 } 96 for(i = 1 ; i <= n ; i++) 97 for(j = 1; j <= n ; j++) 98 { 99 scanf("%d",&a); 100 w[i][a]-=j; 101 } 102 double s = km(); 103 printf("Data Set %d, Best average difference: %.6f ",kk,s-1); 104 dfs(1,0); 105 puts(""); 106 } 107 return 0; 108 }