Polyomino Composer

A polyomino is a plane geometric figure formed by joining one or more equal squares edge to edge.
- Wikipedia

Given a large polyomino and a small polyomino, your task is to determine whether you can compose the large one with two copies of the small one. The polyominoes can be translated, but not flipped or rotated. The two pieces should not overlap. The leftmost picture below is a correct way of composing the large polyomino, but the right two pictures are not. In the middle picture, one of the pieces was rotated. In the rightmost picture, both pieces are exactly identical, but they're both rotated from the original piece (shown in the lower-right part of the picture).

Input

There will be at most 20 test cases. Each test case begins with two integers n and m ( 1

10) in a single line. The next n lines describe the large polyomino. Each of these lines contains exactly n characters in `*',`.'. A `*' indicates an existing square, and a `.' indicates an empty square. The next m lines describe the small polyomino, in the same format. These characters are guaranteed to form valid polyominoes (note that a polyomino contains at least one existing square). The input terminates with n = m = 0, which should not be processed.

Output

For each case, print `1' if the corresponding composing is possible, print `0' otherwise.

Sample Input

4 3 .**. **** .**. .... **. .** ... 3 3 *** *.* *** *.. *.. **. 4 2 **** .... .... .... *. *. 0 0

Sample Output

1 0 0

The Seventh Hunan Collegiate Programming Contest
Problemsetter: Rujia Liu, Special Thanks: Yiming Li & Jane Alam Jan

目描述：

不旋及翻，一形有有可能被小形拼出，而且不互相重。

目解法：

由於定的小示都是正方形，且是通，但是拼的候可能有多的小示超界。

先抓一小示的非空，然後去大示的其中一，接著去它。

接著再抓一被使用的，再第二次，中查是否有碰撞。

#include <stdio.h>
#include <string.h>
char g1[50][50], g2[50][50];
int used[15][15], used2[15][15], err;
int n, m;
void dfs(int x, int y, int vx, int vy) {
    if(x < 0 || y < 0 || x >= m || y >= m)
        return;
    if(err || g2[x][y] == '.' || used2[x][y])
        return;
    if(vx < 0 || vy < 0 || vx >= n || vy >= n || g1[vx][vy] == '.' || used[vx][vy] == 1)
        err = 1;
    used2[x][y] = 1;
    used[vx][vy] = 1;
    dfs(x+1, y, vx+1, vy);
    dfs(x-1, y, vx-1, vy);
    dfs(x, y+1, vx, vy+1);
    dfs(x, y-1, vx, vy-1);
}
int main() {
    int i, j, k, p, q, r, a, b, c;
    while(scanf("%d %d", &n, &m) == 2) {
        if(n == 0 && m == 0)
            break;
        for(i = 0; i < n; i++)
            scanf("%s", &g1[i]);
        for(i = 0; i < m; i++)
            scanf("%s", &g2[i]);
        int px, py;
        for(i = 0; i < m; i++) {
            for(j = 0; j < m; j++) {
                if(g2[i][j] == '*') {
                    px = i, py = j;
                    i = m, j = m;
                }
            }
        }
        int ret = 0;
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++) {
                if(g1[i][j] == '*') {
                    memset(used, 0, sizeof(used));
                    memset(used2, 0, sizeof(used2));
                    err = 0;
                    dfs(px, py, i, j);
                    if(err) continue;
                    for(p = 0; p < n; p++) {
                        for(q = 0; q < n; q++) {
                            if(g1[p][q] == '.' || used[p][q])
                                continue;
                            memset(used2, 0, sizeof(used2));
                            err = 0;
                            dfs(px, py, p, q);
                            if(err) continue;
                            ret = 1;
                            p = n, q = n, i = n, j = n;
                        }
                    }
                }
            }
        }
        printf("%d\n", ret);
    }
    return 0;
}