zoj 1008 Gnome Tetravex

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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

        这道题做了很久，还是没有做出来，陷入了死胡同吧，网上好多人都觉得很简单，这就是差距吧。在网上看到的方法：把不同的方块存在不同的数组中，相同的就存在同一个数组中，并记录个数。然后遍历各个点。还有就是注意格式问题，每两个之间有一个空行，最后那个样例没有空行。

#include <stdio.h>
#include <string.h>

int map[30][4];
int counting[30], mark[30];

int DFS(int num, int n, int cur)
{
    if(num == n*n)
        return 1;
    for(int i = 0; i<cur; i++)
    {
        if(!counting[i])   //如果用完就跳过
            continue;
        if(num%n != 0)   //把不是第一列的来判断
        {
            if(map[i][3] != map[ mark[num-1] ][1])
                continue;
        }
        if(num/n != 0)   //把不是第一行的来判断
        {
            if(map[i][0] != map[ mark[num-n] ][2])
                continue;
        }
        mark[num] = i;
        counting[i]--;
        if(DFS(num+1, n, cur))
            return 1;
        else
            counting[i]++;
    }

    return 0;
}

int main()
{
    int T = 1;
    int n, up, down, left, right, cur;
    while(scanf("%d", &n)!=EOF && n)
    {
        cur = 0;     //初始化
        memset(map, 0, sizeof(map));
        memset(mark, 0, sizeof(mark));
        memset(counting, 0, sizeof(counting));

        for(int i = 0, j; i<n*n; i++)
        {
            scanf("%d%d%d%d", &up, &right, &down, &left);
            for(j = 0; j<cur; j++)
            {
                if(up==map[j][0] && right==map[j][1] && down==map[j][2] && left==map[j][3])  //把相同的存入同一个数组，记录个数
                {
                    counting[j]++;
                    break;
                }
            }
            if(j == cur)   //把不同的存入数组
            {
                map[j][0] = up;    map[j][1] = right;
                map[j][2] = down;  map[j][3] = left;
                counting[j]++;
                cur++;
            }
        }

        if(T > 1)   //用来处理格式问题
            printf("\n");
        if(DFS(0, n, cur))
            printf("Game %d: Possible\n", T++);
        else
            printf("Game %d: Impossible\n", T++);
    }

    return 0;
}

zoj 1008 Gnome Tetravex,布布扣,bubuko.com

zoj 1008 Gnome Tetravex

标签：des   style   blog   http   color   os   io   strong   

原文地址：http://www.cnblogs.com/fengxmx/p/3910609.html