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。

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4th IIUC Inter-University Programming Contest, 2005
D	Largest Square
	Input: standard input Output: standard output
	Problemsetter: Tanveer Ahsan

Given a rectangular grid of characters you have to find out the length of a side of the largest square such that all the characters of the square are same and the center [intersecting point of the two diagonals] of the square is at location (r, c). The height and width of the grid is M and N respectively. Upper left corner and lower right corner of the grid will be denoted by (0, 0) and (M-1, N-1) respectively. Consider the grid of characters given below. Given the location (1, 2) the length of a side of the largest square is 3.

abbbaaaaaa
abbbaaaaaa
abbbaaaaaa
aaaaaaaaaa
aaaaaaaaaa
aaccaaaaaa
aaccaaaaaa

Input

The input starts with a line containing a single integer T (< 21). This is followed by T test cases. The first line of each of them will contain three integers M, N and Q (< 21) separated by a space where M, N denotes the dimension of the grid. Next follows M lines each containing N characters. Finally, there will be Q lines each containing two integers r and c. The value of M and N will be at most 100.

Output

For each test case in the input produce Q+1 lines of output. In the first line print the value of M, N and Q in that order separated by single space. In the next Q lines, output the length of a side of the largest square in the corresponding grid for each (r, c) pair in the input.

Sample Input	Output for Sample Input
1 7 10 4 abbbaaaaaa abbbaaaaaa abbbaaaaaa aaaaaaaaaa aaaaaaaaaa aaccaaaaaa aaccaaaaaa 1 2 2 4 4 6 5 2	7 10 4 3 1 5 1

---

#include <stdio.h>
#include <stdlib.h>

int main() {
    int T;
    char map[102][102];
    int i, M, N, Q, x, y;
    scanf("%d", &T);
    while(T--) {
        scanf("%d %d %d", &M, &N, &Q);
        getchar();
        for(i = 0; i < M; i++)
            gets(map[i]);
        printf("%d %d %d\n", M, N, Q);

        while(Q--) {
            scanf("%d %d", &x, &y);
            int ans = 1;
            int i, a, b;
            for(i = 0; i <= M || i <= N; i++) {
                int flag = 0;
                for(a = x-i; a <= x+i; a++) {
                    for(b = y-i; b <= y+i; b++) {
                        if(a < 0 || b < 0 || a >= M || b >= N) {
                            flag = 1;break;
                        }
                        if(map[a][b] != map[x][y])
                            flag = 1;
                    }
                }
                if(flag == 0) {
                    if(ans < 2*i+1)
                        ans = i*2+1;
                } else
                    break;
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}