2015
09-17

# Fishhead’s Little Game

There is a 3 by 3 grid and each vertex is assigned a number.

It looks like JiuGongGe, but they are different, for we are not going to fill the cell but the edge. For instance,

adding edge 6 �C> 10

The rule of this game is that each player takes turns to add an edge. You will get one point if the edge you just added, together with edges already added before, forms a new square (only square of size 1 is considered). Of course, you get two points if that edge forms two squares. Notice that an edge can be added only once.

forming two squares to get two points

Tom200 and Jerry404 is playing this little game, and have played n rounds when Fishhead comes in. Fishhead wants to know who will be the winner. Can you help him? Assume that Tom200 and Jerry404 are clever enough to make optimal decisions in each round. Every Game starts from Tom200.

The first line of the input contains a single integer T (T <= 100), the number of test cases.
For each case, the first line contains an integers n (12 <= n <= 24), which means they have taken total n rounds in turn. Next n lines each contains two integers a, b (a, b <= 16) representing the two endpoints of the edge.

The first line of the input contains a single integer T (T <= 100), the number of test cases.
For each case, the first line contains an integers n (12 <= n <= 24), which means they have taken total n rounds in turn. Next n lines each contains two integers a, b (a, b <= 16) representing the two endpoints of the edge.

1
15
1 2
1 5
2 6
5 9
6 10
9 10
5 6
2 3
3 7
7 11
10 11
3 4
6 7
7 8
4 8

Case #1: Tom200
Hint
In case 1, Tom200 gets two points when she add edge 5 -> 6, two points in edge 6 -> 7, one point in 4 -> 8.


#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
int dp[10001];
int visit[10001];
int vis[25];
int oth[25];
int ts;
int num(int vist[])
{
int i;
int sum=0;
for(i=1;i<=9;i++)
{
if(vist[i]&&vist[i+3]&&vist[i+12+(i-1)/3]&&vist[i+13+(i-1)/3])sum++;
}
return sum;
}
int dos(int x)
{
int sum=0,i;
int vist[26];
for(i=0;i<=24;i++)vist[i]=0;
for(i=1;i<=ts;i++)
{
if(x&(1<<(i-1)))
{
vist[oth[i-1]]=1;
}
}
for(i=1;i<=24;i++)
{
if(vis[i])vist[i]=1;
}
sum=num(vist);
return sum;
}
int dfs(int x)
{
if(visit[x]!=-1)return visit[x];
int ans=0;
for(int i=1;i<=ts;i++)
{
if(!(x&(1<<(i-1))))
{
int y;
y=x+(1<<(i-1));
int ss;
ss=dfs(y);
ans=max(9-dp[x]-ss,ans);
}
}
visit[x]=ans;
return ans;
}
int main()
{
int T,i,n,a,b;
int cas;
int anum,bnum,s;
cas=0;
scanf("%d",&T);
while(T--)
{
cas++;
anum=bnum=0;
s=0;
scanf("%d",&n);
memset(vis,0,sizeof(vis));
memset(visit,-1,sizeof(visit));
for(i=0;i<n;i++)
{
scanf("%d%d",&a,&b);
if(a>b)swap(a,b);
if(b-a==4)ts=12+a;
else ts=a-a/4;
vis[ts]=1;
if(i%2==0)
{
anum+=num(vis)-s;
}
else bnum+=num(vis)-s;
s=num(vis);
}
ts=0;
for(i=1;i<=24;i++)if(vis[i]==0)oth[ts++]=i;
for(i=0;i<(1<<ts);i++)dp[i]=dos(i);
int have;
have=9-s;
int fs;
fs=dfs(0);
//printf("%d %d %d %d\n",anum,bnum,have,fs);
printf("Case #%d: ",cas);
if(n%2==0)
{
if(anum+fs>bnum+have-fs)cout<<"Tom200"<<endl;
else cout<<"Jerry404"<<endl;
}
else
{
if(anum+have-fs>bnum+fs)cout<<"Tom200"<<endl;
else cout<<"Jerry404"<<endl;
}
}
return 0;
}


1. 回你幼稚园去吧，这本来就是开玩笑的你他娘的当真了显得你很正义吗？我就是翻了后面的评论才来给你屎吃的，你还是不明白吗？没事，多来点脑白金你可能就没事了