Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Sheryl works for a software company in the country of Brada. Her job is to develop a Windows operating system. People in Brada are incredibly conservative. They even never use graphical monitors! So Sheryl's operating system has to run in text mode and windows in that system are formed by characters. Sheryl decides that every window has an ID which is a capital English letter (`A' to `Z'). Because every window had a unique ID, there can't be more than 26 windows at the same time. And as you know, all windows are rectangular.

On the screen of that ugly Windows system, a window's frame is formed by its ID letters. Fig-1 shows that there is only one window on the screen, and that window's ID is `A'. Windows may overlap. Fig-2 shows the situation that window B is on the top of window A. And Fig-3 gives a more complicated overlapping. Of course, if some parts of a window are covered by other windows, you can't see those parts on the screen.

.........................
....AAAAAAAAAAAAA........
....A...........A........
....A...........A........
....A...........A........
....AAAAAAAAAAAAA........
.........................

Fig-1

.........................
....AAAAAAAAAAAAA........
....A...........A........
....A.......BBBBBBBBBB...
....A.......B........B...
....AAAAAAAAB........B...
............BBBBBBBBBB...
.........................

Fig-2

..........................
....AAAAAAAAAAAAA.........
....A...........A.........
....A.......BBBBBBBBBB....
....A.......B........BCCC.
....AAAAAAAAB........B..C.
.......C....BBBBBBBBBB..C.
.......CCCCCCCCCCCCCCCCCC.
..........................

Fig-3

If a window has no parts covered by other windows, we call it a ``top window" (The frame is also considered as a part of a window). Usually, the top windows are the windows that interact with user most frequently. Assigning top windows more CPU time and higher priority will result in better user experiences. Given the screen presented as Figs above, can you tell Sheryl which windows are top windows?

Input 

The input contains several test cases.

Each test case begins with two integers, n and m (1n, m100) , indicating that the screen has n lines, and each line consists of m characters.

The following n lines describe the whole screen you see. Each line contains m characters. For characters which are not on any window frame, we just replace them with `.' .

The input ends with a line of two zeros.

It is guaranteed that:

1)

There is at least one window on the screen.

2)

Any window's frame is at least 3 characters wide and 3 characters high.

3)

No part of any window is outside the screen.

Output 

For each test case, output the IDs of all top windows in a line without blanks and in alphabet order.

Sample Input 

9 26 
..........................
....AAAAAAAAAAAAA.........
....A...........A.........
....A.......BBBBBBBBBB....
....A.......B........BCCC.
....AAAAAAAAB........B..C.
.......C....BBBBBBBBBB..C.
.......CCCCCCCCCCCCCCCCCC.
..........................
7 25
.........................
....DDDDDDDDDDDDD........
....D...........D........
....D...........D........
....D...........D..AAA...
....DDDDDDDDDDDDD..A.A...
...................AAA... 
0 0

Sample Output 

B 
AD

---

抓左上角點，然後把點點都拿出來判斷周圍有沒有異物。

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
char g[105][105];
int dir[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
int n, m, i, j, p;
int check(int x, int y, int c) {
    int tx, ty, i;
    for(i = 0; i < 4; i++) {
        tx = x+dir[i][0], ty = y+dir[i][1];
        if(tx < 0 || ty < 0 || tx >= n || ty >= m)
            continue;
        if(g[tx][ty] != '.' && g[tx][ty] != c)
            return 0;
    }
    return 1;
}
int main() {
    while(scanf("%d %d", &n, &m) == 2) {
        if(n == 0)  break;
        memset(g, 0, sizeof(g));
        for(i = 0; i < n; i++)
            scanf("%s", g[i]);
        int alph[26] = {}, x, y, tx, ty;
        for(i = 0; i < n; i++) {
            for(j = 0; j < m; j++) {
                if(g[i][j] >= 'A' && g[i][j] <= 'Z'
                   && g[i][j] == g[i][j+1] && g[i][j] == g[i+1][j]) { // left upper corner
                    if(g[i+1][j+1] == '.') {
                        queue<int> X, Y;
                        X.push(i+1), Y.push(j+1);
                        int flag = 1; // true
                        char used[105][105] = {};
                        while(!X.empty() && flag) {
                            x = X.front(), X.pop();
                            y = Y.front(), Y.pop();
                            flag &= check(x, y, g[i][j]);
                            for(p = 0; p < 4; p++) {
                                tx = x+dir[p][0], ty = y+dir[p][1];
                                if(tx < 0 || ty < 0 || tx >= n || ty >= m)
                                    continue;
                                if(used[tx][ty] || g[tx][ty] != '.')
                                    continue;
                                used[tx][ty] = 1;
                                X.push(tx), Y.push(ty);
                            }
                        }
                        if(flag)    alph[g[i][j]-'A'] = 1;
                    }
                }
            }
        }
        for(i = 0; i < 26; i++)
            if(alph[i])
                putchar(i+'A');
        puts("");
    }
    return 0;
}