UVA 10189 Minesweeper
Have you ever played Minesweeper? It’s a cute little game which comes within a certain Operating System which name we can’t really remember. Well, the goal of the game is to ?nd where are all the mines within a M ×N ?eld. To help you, the game shows a number in a square which tells you how many mines there are adjacent to that square. For instance, supose the following 4×4 ?eld with 2 mines (which are represented by an ‘*’ character):
*...
....
.*..
....
If we would represent the same ?eld placing the hint numbers described above, we would end up with:
*100
2210
1*10
1110
As you may have already noticed, each square may have at most 8 adjacent squares.
Input
The input will consist of an arbitrary number of ?elds. The ?rst line of each ?eld contains two integers n and m (0 < n,m ≤ 100) which stands for the number of lines and columns of the ?eld respectively. The next n lines contains exactly m characters and represent the ?eld.
Each safe square is represented by an ‘.’ character (without the quotes) and each mine square is represented by an ‘*’ character (also without the quotes). The ?rst ?eld line where n = m = 0 represents the end of input and should not be processed.
Output
For each ?eld, you must print the following message in a line alone:
Field #x:
Where x stands for the number of the ?eld (starting from 1). The next n lines should contain the ?eld with the ‘.’ characters replaced by the number of adjacent mines to that square. There must be an empty line between ?eld outputs.
Sample Input
4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0
Sample Output
Field #1:
*100
2210
1*10
1110
Field #2:
**100
33200
1*100
思路：可以一开始的时候把存field的数组都初始化为'0'，如果输入‘*’，则用'*'取代相应位置上的字符，然后以它为中心的8-邻域中不是‘*’的位置都可以+1，因为一个‘*’只会对它的8-邻域产生影响，如果输入‘.'，则无需操作。
代码：
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<map>
#include<set>
using namespace std;char field[105][105];int main()
{int m, n, t = 0;scanf("%d%d", &m, &n);while (m && n){for (int i = 0; i < m; i++){for (int j = 0; j < n; j++)field[i][j] = '0';}for (int i = 0; i < m; i++){getchar();//过滤回车for (int j = 0; j < n; j++){ char ch = getchar();if (ch == '*'){field[i][j] = '*';if (i - 1 >= 0 && j - 1 >= 0 && field[i - 1][j - 1] != '*')field[i - 1][j - 1]++;if (i - 1 >= 0 && field[i - 1][j] != '*')field[i - 1][j]++;if (i - 1 >= 0 && j + 1 < n && field[i - 1][j + 1] != '*')field[i - 1][j + 1]++;if (j - 1 >= 0 && field[i][j - 1] != '*')field[i][j - 1]++;if (j + 1 < n && field[i][j + 1] != '*')field[i][j + 1]++;if (i + 1 < m && j - 1 >= 0 && field[i + 1][j - 1] != '*')field[i + 1][j - 1]++;if (i + 1 < m && field[i + 1][j] != '*')field[i + 1][j]++;if (i + 1 < m && j + 1 < n && field[i + 1][j + 1] != '*')field[i + 1][j + 1]++;}} }if(t)printf("\n");printf("Field #%d:\n", ++t);for (int i = 0; i < m; i++){for (int j = 0; j < n; j++)printf("%c", field[i][j]);printf("\n");}scanf("%d%d", &m, &n);}return 0;
}