ZOJ1008-Gnome Tstravex（dfs）

Gnome Tetravex

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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

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Source: Asia 2001, Shanghai (Mainland China)

思路：dfs+简单剪枝；简单剪枝：重复的化为一个类型；

代码：

#include <stdio.h>int n;//方格矩阵的大小 int cnt=0;//方格的类型数 int sq[25][4];//方格矩阵 int sum[25];//存放方格的类型int ans[25];//存放解决方案 int cas=0;//测试例计数 int a,b,c,d;//上，右，下，左 bool dfs(int index){if(index==n*n)    return true;//cnt种方格类型，每一个都往index位置放置一下，看看是否符合要求 for(int i=0;i<cnt;i++){//这种类型方格用完 if(sum[i]==0)    continue;//该方格不在最左边 if(index%n!=0)    if(sq[ans[index-1]][1]!=sq[i][3])        continue;//该方格不在最上边 if(index>=n)    if(sq[ans[index-n]][2]!=sq[i][0])        continue;//index位置放置第i种类型方格 ans[index]=i;//该类型方格减1 sum[i]--;if(dfs(index+1))    return true;//恢复该方格的类型数，以便下一次搜索 sum[i]++;}return false; }  int main() { while(scanf("%d",&n)&&n) { int j;cas++; for(int i=0;i<n*n;i++) { scanf("%d%d%d%d",&a,&b,&c,&d); //判断和以前的方格类型是否相同 for(j=0;j<cnt;j++) { if(sq[j][0]==a&&sq[j][1]==b&&sq[j][2]==c&&sq[j][3]==d) {     sum[j]++;     break;     }}//与以前方格类型不相同时 if(j==cnt){sq[j][0]=a;sq[j][1]=b;sq[j][2]=c;sq[j][3]=d;sum[j]=1; cnt++;}}    if(cas>1)  printf("\n");//测试例之间的回车 if(dfs(0))    printf("Game %d: Possible\n",cas);else    printf("Game %d: Impossible\n",cas);}return 0; }