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 A
Fill the Square
Input: Standard Input

Output: Standard Output

 

In this problem, you have to draw a square using uppercase English Alphabets.

To be more precise, you will be given a square grid with some empty blocks and others already filled for you with some letters to make your task easier. You have to insert characters in every empty cell so that the whole grid is filled with alphabets. In doing so you have to meet the following rules:

 

1. Make sure no adjacent cells contain the same letter; two cells are adjacent if they share a common edge.  
2. There could be many ways to fill the grid. You have to ensure you make the lexicographically smallest one. Here, two grids are checked in row major order when comparing lexicographically.

 

Input
The first line of input will contain an integer that will determine the number of test cases. Each case starts with an integer n( n<=10 ), that represents the dimension of the grid. The next n lines will contain n characters each. Every cell of the grid is either a ‘.’ or a letter from [A, Z]. Here a ‘.’ Represents an empty cell.

 

Output

For each case, first output Case #: ( # replaced by case number ) and in the next n lines output the input matrix with the empty cells filled heeding the rules above.

Sample Input                       Output for Sample Input  

2 3 ... ... ... 3 ... A.. ...

Case 1: ABA BAB ABA Case 2: BAB ABA BAB

Problemsetter: Sohel Hafiz

---

#include <stdio.h>

int main() {
    int cases = 0;
    int t, n ,i, j, k, l;
    char map[20][20];
    scanf("%d", &t);
    while(t--) {
        printf("Case %d:\n", ++cases);
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            scanf("%s", map[i]);
        int d[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++) {
                if(map[i][j] == '.') {
                    for(k = 'A'; k <= 'Z'; k++) {
                        map[i][j] = k;
                        for(l = 0; l < 4; l++) {
                            if(d[l][0]+i < 0 || d[l][0]+i >= n ||
                               d[l][1]+j < 0 || d[l][1]+j >= n)
                                continue;
                            if(map[i][j] == map[i+d[l][0]][j+d[l][1]])
                                break;
                        }
                        if(l == 4)  break;
                    }
                }
            }
        }
        for(i = 0; i < n; i++)
            puts(map[i]);
    }
    return 0;
}