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

Problem D: You want what filled?

Now that Ryan and Larry infuriated the bear, they're now forced to do menial tasks for food, one such task is filling in potholes on this deserted island. But of course, it's never quite that easy, as the bear forces them to fill in the biggest hole first. Since Ryan and Larry are still lazy (very little has changed, you see), can you tell them the order to fill in the holes?

Input

The first line will contain two numbers x and y, followed by x lines of y characters each. (x and y are less than 50). The holes will be represented by the uppercase characters A to Z, and regular land will be represented by ".". There will be no other characters in the map. Input will be terminated by having 0 0 as input.

Output

For each map, output the problem number (as shown below), then output the hole represented by the character, and the number of space the hole takes up, sorted by the size of the hole, break ties by sorting the characters in alphabetical order, as shown in the sample output on a separate line as shown below:

Sample Input

5 5
..AAA
E.BBB
..AA.
CC.DD
CC.D.
5 5
..AAA
E.BBB
..AA.
CC.DD
CC.D.
0 0

Sample Output

Problem 1:
C 4
A 3
B 3
D 3
A 2
E 1
Problem 2:
C 4
A 3
B 3
D 3
A 2
E 1

---

#include <stdio.h>
#include <stdlib.h>
typedef struct {
    int v;
    char c;
} Out;
int cmp(const void *i, const void *j) {
    Out *a, *b;
    a = (Out *)i, b = (Out *)j;
    if(a->v != b->v)
        return b->v - a->v;
    return a->c - b->c;
}
int main() {
    int n, m, test = 0, i, j, k, l;
    char map[51][51];
    Out ans[2500];
    while(scanf("%d %d", &n, &m) == 2) {
        if(n == 0 && m == 0)
            break;
        for(i = 0; i < n; i++)
            scanf("%s", map[i]);
        int used[51][51] = {}, D[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
        int Q[2500][2], Qt, idx = 0, x, y, xx, yy;
        for(i = 0; i < n; i++) {
            for(j = 0; j < m; j++) {
                if(used[i][j] == 0 && map[i][j] != '.') {
                    Qt = 0, Q[0][0] = i, Q[0][1] = j, used[i][j] = 1;
                    for(k = 0; k <= Qt; k++) {
                        x = Q[k][0], y = Q[k][1];
                        for(l = 0; l < 4; l++) {
                            xx = x+D[l][0], yy = y+D[l][1];
                            if(xx < 0 || yy < 0 || xx >= n || yy >= m)
                                continue;
                            if(used[xx][yy] == 0 && map[i][j] == map[xx][yy]) {
                                Qt++;
                                Q[Qt][0] = xx, Q[Qt][1] = yy;
                                used[xx][yy] = 1;
                            }
                        }
                    }
                    ans[idx].v = Qt+1, ans[idx].c = map[i][j];
                    idx++;
                }
            }
        }
        qsort(ans, idx, sizeof(Out), cmp);
        printf("Problem %d:\n", ++test);
        for(i = 0; i < idx; i++)
            printf("%c %d\n", ans[i].c, ans[i].v);
    }
    return 0;
}