Description

Consider the following 5 picture frames placed on an 9 x 8 array. 

........ ........ ........ ........ .CCC....

EEEEEE.. ........ ........ ..BBBB.. .C.C....

E....E.. DDDDDD.. ........ ..B..B.. .C.C....

E....E.. D....D.. ........ ..B..B.. .CCC....

E....E.. D....D.. ....AAAA ..B..B.. ........

E....E.. D....D.. ....A..A ..BBBB.. ........

E....E.. DDDDDD.. ....A..A ........ ........

E....E.. ........ ....AAAA ........ ........

EEEEEE.. ........ ........ ........ ........

    1        2        3        4        5

Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below.

Viewing the stack of 5 frames we see the following. 

.CCC....

ECBCBB..

DCBCDB..

DCCC.B..

D.B.ABAA

D.BBBB.A

DDDDAD.A

E...AAAA

EEEEEE..



In what order are the frames stacked from bottom to top? The answer is EDABC.

Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules:

1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters.

2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides.

3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter.

Input

Each input block contains the height, h (h<=30) on the first line and the width w (w<=30) on the second. A picture of the stacked frames is then given as h strings with w characters each.
Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially.

Output

Write the solution to the standard output. Give the letters of the frames in the order they were stacked from bottom to top. If there are multiple possibilities for an ordering, list all such possibilities in alphabetical order, each one on a separate line. There will always be at least one legal ordering for each input block. List the output for all blocks in the input sequentially, without any blank lines (not even between blocks).

Sample Input

9
8
.CCC....
ECBCBB..
DCBCDB..
DCCC.B..
D.B.ABAA
D.BBBB.A
DDDDAD.A
E...AAAA
EEEEEE..

Sample Output

EDABC
题目：要求对给出的矩阵，求出它们的叠放顺序。

解法：对每种字母，求出相同字母的最小区域。然后用递归去枚举每一种情况。原来是打算用深搜，但后来发现不可以。
            AABBA
            AXXXA
            如果用深搜，对每一个搜索过的字母标记为‘X’，表示此点可以被连通。则在上面的情况，先搜A的话，也可以搜出连通，但这实际是不行的。

代码如下：

#include<stdio.h>
#include<string.h>
#define max 31
int o, H[26], I, M[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}, T[26];
char L[27], F[10000][27];
struct N
{
    int x, y;
    char ch;
}P[26][100];
struct CC
{
    int x1, y1, x2, y2, x, y;
    char ch;
}N[26];
int cmp(const void *a, const void *b)
{
    return strcmp((char*)a, (char*)b);
}
int Dfs(int g, int n, int m, char s[][max], int depth)
{
    int i, j, k, x, y;
    x = N[g].x;
    y = N[g].y;    
    for (i = N[g].x1, k = N[g].x2, j = N[g].y1; j <= N[g].y2; j++)
    {
        if (s[i][j] != N[g].ch && s[i][j] != 'X')
        {
            return 0;
        }
        else
        {
            P[depth][I].x = i;
            P[depth][I].y = j;
            P[depth][I++].ch = s[i][j];
        }
        if (s[k][j] != N[g].ch && s[k][j] != 'X')
        {
            return 0;
        }
        else
        {
            P[depth][I].x = k;
            P[depth][I].y = j;
            P[depth][I++].ch = s[k][j];
        }
    }
    for (j = N[g].y1, k = N[g].y2, i = N[g].x1; i <= N[g].x2; i++)
    {
        if (s[i][j] != N[g].ch && s[i][j] != 'X')
        {
            return 0;
        }
        else
        {
            P[depth][I].x = i;
            P[depth][I].y = j;
            P[depth][I++].ch = s[i][j];
        }
        if (s[i][k] != N[g].ch && s[i][k] != 'X')
        {
            return 0;
        }
        else
        {
            P[depth][I].x = i;
            P[depth][I].y = k;
            P[depth][I++].ch = s[i][k];    
        }
    }
    return 1;    
}
void Cu(int g, int n, int m, int depth, char s[][max], int num)
{
    int i, j, t;
    I = 0;
    t = Dfs(g, n, m, s, depth);
    H[depth] = I;
    if (t)
    { 
        
        L[depth] = N[g].ch;
        if (depth + 1 == num)
        {
            for (i = depth; i >= 0; i--)
            {
                F[o][depth - i] = L[i];
            }
            F[o++][depth + 1] = '\0';
        }
        else
        {
            for (j = 0; j < H[depth]; j++)
            {
                s[P[depth][j].x][P[depth][j].y] = 'X';
            }
            for (i = 0; i < num; i++)
            {
                if (!T[i])
                {
                    T[i] = 1;                    
                    Cu(i, n, m, depth + 1, s, num);                    
                    T[i] = 0;
                }
            }
            for (j = 0; j < H[depth]; j++)
            {
                s[P[depth][j].x][P[depth][j].y] = P[depth][j].ch;
            }
        }
    }
}
                
int main()
{
    char s[max][max], t1[26], ch, str[max][max];
    int i, j, k, a, b, n, m, t[26], x1, y1, x2, y2, l, h;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        getchar();
        for (i = 0; i < 26; i++)
        {
            t[i] = 0;
        }
        for (i = 0; i < n; i++)
        {
            scanf("%s", &s[i]);
        }
        for (i = 0, k = 0; i < n; i++)
        {
            for (j = 0; j < m; j++)
            {
                if (s[i][j] != '.' && t[s[i][j] - 'A' ] == 0)
                {
                    t[s[i][j] - 'A'] = 1;
                    x1 = 100;
                    y1 = 100;
                    x2 = 0;
                    y2 = 0;
                    for (l = 0; l < n; l++)
                    {
                        for (h = 0; h < m; h++)
                        {
                            if (s[l][h] == s[i][j])
                            {
                                if (l < x1)
                                {
                                    x1 = l;
                                }
                                if (h < y1)
                                {
                                    y1 = h;
                                }
                                if (x2 < l)
                                {
                                    x2 = l;
                                }
                                if (y2 < h)
                                {
                                    y2 = h;
                                }
                            }
                        }
                    }
                    N[k].x = i;
                    N[k].y = j;
                    N[k].x1 = x1;
                    N[k].x2 = x2;
                    N[k].y1 = y1;
                    N[k].ch = s[i][j];
                    N[k++].y2 = y2;                                    
                }
            }
        }
        o = 0;
        for (i = 0; i < k; i++)
        {
            for (j = 0; j < n; j++)
            {
                strcpy(str[j], s[j]);
            }
            T[i] = 1;
            Cu(i, n, m, 0, str, k);
            T[i] = 0;
        }
        qsort(F, o, sizeof(char) * 27, cmp);
        for (i = 0; i < o; i++)puts(F[i]);
    }
    return 0;
}