思路：状态最多有2^12，采用记忆化搜索！！
代码如下：
1 #include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<iomanip> 5 #include<cmath> 6 #include<cstring> 7 #include<vector> 8 #define inf 1<<30 9 using namespace std; 10 int edge[][2]={{1,2},{2,3},{3,4},{1,5},{2,6},{3,7},{4,8},{5,6},{6,7},{7,8},{5,9}, 11 {6,10},{7,11},{8,12},{9,10},{10,11},{11,12},{9,13},{10,14},{11,15},{12,16},{13,14}, 12 {14,15},{15,16}}; 13 int tri[][4]={{0,3,4,7},{1,4,5,8},{2,5,6,9},{7,10,11,14},{8,11,12,15}, 14 {9,12,13,16},{14,17,18,21},{15,18,19,22},{16,19,20,23}}; 15 int p[25],index[25],cnt; 16 int dp[1<<12]; 17 int cal(int state,int e) 18 { 19 int ans=0,t=0; 20 for(int i=0;i<9;i++){ 21 bool flag=0; 22 for(int j=0;j<4;j++) 23 if(e==tri[i][j]){ 24 flag=true; 25 break; 26 } 27 if(flag){ 28 t++; 29 for(int j=0;j<4;j++){ 30 if(!(p[tri[i][j]]&state||e==tri[i][j])){ 31 ans--;break; 32 } 33 } 34 ans++; 35 } 36 if(t==2) break; 37 } 38 return ans; 39 } 40 int dfs(int state,int now) 41 { 42 if(dp[now]!=-inf) return dp[now]; 43 int ans=-inf; 44 for(int i=0;i<cnt;i++){ 45 if(!(now&p[i])){ 46 int tt=cal(state,index[i]); 47 tt-=dfs(state|p[index[i]],now|p[i]); 48 ans=max(ans,tt); 49 } 50 } 51 return dp[now]=ans; 52 } 53 int main() 54 { 55 int t,u,v,ca=0,m; 56 bool vis[25]; 57 p[0]=1; 58 for(int i=1;i<=24;i++) p[i]=p[i-1]*2; 59 scanf("%d",&t); 60 while(t--){ 61 scanf("%d",&m); 62 int m0=0,m1=0,tt=0,state=0,side=0; 63 memset(vis,0,sizeof(vis)); 64 while(m--){ 65 scanf("%d%d",&u,&v); 66 if(u>v) swap(u,v); 67 for(int i=0;i<24;i++) 68 if(edge[i][0]==u&&edge[i][1]==v){ 69 tt=cal(state,i); 70 vis[i]=1; 71 state|=p[i]; 72 side==0?m0+=tt:m1+=tt; 73 side++; 74 side%=2; 75 break; 76 } 77 } 78 cnt=0; 79 for(int i=0;i<24;i++){ 80 if(!vis[i]){ 81 index[cnt++]=i; 82 } 83 } 84 for(int i=0;i<(1<<cnt);i++) dp[i]=-inf; 85 dp[(1<<cnt)-1]=0; 86 int ans=m0-m1; 87 if(side==0) ans+=dfs(state,0); 88 else ans-=dfs(state,0); 89 printf("Case #%d: %s ",++ca,ans>0?"Tom200":"Jerry404"); 90 } 91 return 0; 92 }