http://acm.jlu.edu.cn/joj/showproblem.php?pid=2630&off=2600
此题是2008年杭州赛区的第一道题，题意其实就是同构图的判定问题。根据数据量，枚举是个可行的办法。代码如下：
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1 #include <stdio.h> 2 #include <memory.h> 3 #define MAX 1<<30 4 int ca=1,n,ma,mb,ia,ib,da,db; 5 int mincost,vis[9],map_v[9],edge_a[9][9],edge_b[9][9]; 6 int min(int a,int b,int c,int d) 7 { 8 if(a<=b&&a<=c&&a<=d)return a; 9 if(b<=a&&b<=c&&b<=d)return b; 10 if(c<=a&&c<=b&&c<=d)return c; 11 return d; 12 } 13 void dfs(int step) 14 { 15 if(mincost==0)return;//可能会提高速度的剪枝 16 if(step==n) 17 { 18 int i,j,dif_ma=ma,dif_mb=mb; 19 for(i=0;i<n;i++) 20 { 21 for(j=i+1;j<n;j++) 22 { 23 if(edge_b[i][j]&&edge_a[map_v[i]][map_v[j]]) 24 { 25 dif_ma--; 26 dif_mb--; 27 } 28 } 29 } 30 int cur_mincost = min(dif_mb*ia+dif_ma*ib,dif_ma*da+dif_mb*db,dif_ma*da+dif_mb*ia,dif_mb*db+dif_ma*ib); 31 if(cur_mincost<mincost)mincost = cur_mincost; 32 return; 33 } 34 int j; 35 for(j=0;j<n;j++) 36 { 37 if(!vis[j]) 38 { 39 vis[j] = 1; 40 map_v[step] = j; 41 dfs(step+1); 42 vis[j] = 0; 43 } 44 } 45 } 46 int main() 47 { 48 int i; 49 while(scanf("%d%d%d",&n,&ma,&mb),n+ma+mb) 50 { 51 memset(edge_a,0,sizeof(edge_a)); 52 memset(edge_b,0,sizeof(edge_b)); 53 scanf("%d%d%d%d",&ia,&ib,&da,&db); 54 for(i=0;i<ma;i++) 55 { 56 int a,b; 57 scanf("%d%d",&a,&b); 58 edge_a[a][b]=edge_a[b][a] = 1; 59 60 } 61 for(i=0;i<mb;i++) 62 { 63 int a,b; 64 scanf("%d%d",&a,&b); 65 edge_b[a][b]=edge_b[b][a] = 1; 66 } 67 mincost = MAX; 68 memset(vis,0,sizeof(vis)); 69 for(i=0;i<n;i++)map_v[i]=i; 70 dfs(0); 71 printf("Case #%d: %d\n",ca++,mincost); 72 } 73 return 0; 74 }
另外，C++ STL有生成排列数的库函数，效率比一般自己编写的要高，试过之后，速度果然有提高。代码如下：
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1 #include <iostream> 2 #include<algorithm> 3 using namespace std; 4 #define MAX 1<<30 5 int ca=1,n,ma,mb,ia,ib,da,db; 6 int mincost,map_v[9],edge_a[9][9],edge_b[9][9]; 7 int min(int a,int b,int c,int d) 8 { 9 if(a<=b&&a<=c&&a<=d)return a; 10 if(b<=a&&b<=c&&b<=d)return b; 11 if(c<=a&&c<=b&&c<=d)return c; 12 return d; 13 } 14 int main() 15 { 16 int i; 17 while(scanf("%d%d%d",&n,&ma,&mb),n+ma+mb) 18 { 19 memset(edge_a,0,sizeof(edge_a)); 20 memset(edge_b,0,sizeof(edge_b)); 21 scanf("%d%d%d%d",&ia,&ib,&da,&db); 22 for(i=0;i<ma;i++) 23 { 24 int a,b; 25 scanf("%d%d",&a,&b); 26 edge_a[a][b]=edge_a[b][a] = 1; 27 } 28 for(i=0;i<mb;i++) 29 { 30 int a,b; 31 scanf("%d%d",&a,&b); 32 edge_b[a][b]=edge_b[b][a] = 1; 33 } 34 mincost = MAX; 35 int map_v[8]={0,1,2,3,4,5,6,7}; 36 do 37 { 38 int i,j,dif_ma=ma,dif_mb=mb; 39 for(i=0;i<n;i++) 40 { 41 for(j=i+1;j<n;j++) 42 { 43 if(edge_b[i][j]&&edge_a[map_v[i]][map_v[j]]) 44 { 45 dif_ma--; 46 dif_mb--; 47 } 48 } 49 } 50 int cur_mincost = min(dif_mb*ia+dif_ma*ib,dif_ma*da+dif_mb*db,dif_ma*da+dif_mb*ia,dif_mb*db+dif_ma*ib); 51 if(cur_mincost<mincost)mincost = cur_mincost; 52 }while (next_permutation(map_v,map_v+n)); 53 printf("Case #%d: %d\n",ca++,mincost); 54 } 55 }