ZOJ 1008 Gnome Tetravex（DFS）

Gnome Tetravex

Time Limit: 10 Seconds
Memory Limit: 32768 KB

Hart is engaged in playing an interesting game, Gnome Tetravex, these days. In the game, at the beginning, the player is given n*n squares. Each square is divided into four triangles marked four numbers (range from 0 to 9). In a square, the triangles are
the left triangle, the top triangle, the right triangle and the bottom triangle. For example, Fig. 1 shows the initial state of 2*2 squares.



Fig. 1 The initial state with 2*2 squares
The player is required to move the squares to the termination state. In the termination state, any two adjoining squares should make the adjacent triangle marked with the same number. Fig. 2 shows one of the termination states of the above example.



Fig. 2 One termination state of the above example
It seems the game is not so hard. But indeed, Hart is not accomplished in the game. He can finish the easiest game successfully. When facing with a more complex game, he can find no way out.

One day, when Hart was playing a very complex game, he cried out, "The computer is making a goose of me. It's impossible to solve it." To such a poor player, the best way to help him is to tell him whether the game could be solved. If he is told the game is
unsolvable, he needn't waste so much time on it.

Input

The input file consists of several game cases. The first line of each game case contains one integer n, 0 <= n <= 5, indicating the size of the game.

The following n*n lines describe the marking number of these triangles. Each line consists of four integers, which in order represent the top triangle, the right triangle, the bottom triangle and the left triangle of one square.

After the last game case, the integer 0 indicates the termination of the input data set.

Output

You should make the decision whether the game case could be solved. For each game case, print the game number, a colon, and a white space, then display your judgment. If the game is solvable, print the string "Possible". Otherwise, please print "Impossible"
to indicate that there's no way to solve the problem.

Print a blank line between each game case.

Note: Any unwanted blank lines or white spaces are unacceptable.

Sample Input

2

5 9 1 4

4 4 5 6

6 8 5 4

0 4 4 3

2

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

0

Output for the Sample Input

Game 1: Possible

Game 2: Impossible

思路：先存不同方块的个数，然后挨个放，能放满就说明可能。否则就不可能。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int mm=33;
class node
{
public:
int u,r,d,l,n;
node(){n=0;}
}f[mm];
int rs[mm];
int n,u,r,d,l,pos;
bool dfs(int x)
{
if(x==n*n)return 1;
for(int i=0;i<pos;i++)
{
if(x/n)
{ ///cout<<f[i].u<<" "<<f[i-n].d<<endl;
if(f[i].u^f[rs[x-n]].d)continue;///符合上下
}
if(x%n)
{///cout<<f[i].l<<" "<<f[i-1].r<<endl;
if(f[i].l^f[rs[x-1]].r)continue;///符合左右
}
if(f[i].n==0)continue;
f[i].n--;
rs[x]=i;
if(dfs(x+1))return 1;
f[i].n++;
}
return 0;
}
int main()
{ int cas=0;
while(cin>>n)
{  pos=0;
memset(f,0,sizeof(f));
if(n==0)break;
memset(rs,0,sizeof(rs));
for(int i=0;i<n*n;i++)
{
cin>>u>>r>>d>>l;
int j;
for(j=0;j<pos;j++)
if(u==f[j].u&&r==f[j].r&&d==f[j].d&&l==f[j].l)///存不同的方块
{
f[j].n++;break;
}
if(j==pos){f[pos].u=u;f[pos].r=r;f[pos].d=d;f[pos].l=l;f[pos].n++;pos++;}
}
if(cas)cout<<"\n";
++cas;
cout<<"Game "<<cas<<": ";
if(dfs(0))
cout<<"Possible\n";
else cout<<"Impossible\n";
}
}