HDU 2514 Another Eight Puzzle

#include <iostream>
using namespace std;
int judge[8][8] = {{3,1,2,3},{4,0,2,4,5},{6,0,1,3,4,5,6},{4,0,2,5,6},{4,1,2,5,7},{6,1,2,3,4,6,7},{4,2,3,5,7},{3,4,5,6}};
//judge[i][0] = 第i列相连点的数量,judge[i][j](j>0) = 相连点.图的数组结构
int num = 0;     //记录解的数量
bool vis[9];     // vis记录8个数中被用过的数
int a[8],fil[8],seq[8];; // fil记录有多少个待填点
int n;
void dfs(int t)
{
if(num > 1) return;
if(t == n)
{
num ++ ;
for(int i=0;i < 8;i++)
{
seq[i] = a[i];
}
return;
}

for(int i = 1;i <= 8;i++) //枚举8个数
{
if(!vis[i])
{
for(int j=1;j < judge[fil[t]][0];j++) //枚举相连点
{
if(abs(a[judge[fil[t]][j]] - i) == 1 && a[judge[fil[t]][j]] != 0) //若连续则跳过
{
break;
}
}
if(j < judge[fil[t]][0]) continue;
//若非正常退出,则出现连续break跳出则不满足条件
//否则,进行深搜
vis[i] = 1;
a[fil[t]] = i;
dfs(t+1);
if(num > 1) return;
vis[i] = 0;
a[fil[t]] = 0;

}
}
}

int main(int argc, const char *argv[])
{
int T;
//    freopen("input.txt","r",stdin);
cin>>T;
for(int k = 1;k <= T;k++)
{
n = 0;
memset(vis,0,sizeof(vis));
for(int i=0;i < 8;i++)
{
cin>>a[i];
if(a[i]) vis[a[i]] = 1;
else if(!a[i])  //若为0.fil记录位置
{
fil[n++] = i;
}
}
num = 0;
dfs(0);
cout<<"Case "<<k<<":"<<" ";
if(num == 1)
{
cout<<seq[0];
for(int i=1;i < 8;i++)
{
cout<<" "<<seq[i];
}
cout<<endl;
}
else if(num > 1)
cout<<"Not unique"<<endl;
else cout<<"No answer"<<endl;
}

return 0;
}