| Problem B: Minesweeper |
The Problem
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 find where are all the mines within a MxN field. 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 4x4 field with 2 mines (which are represented by an * character):
*... .... .*.. ....
If we would represent the same field 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.
The Input
The input will consist of an arbitrary number of fields. The first line of each field contains two integers n and m (0 < n,m <= 100) which stands for the number of lines and columns of the field respectively. The next n lines contains exactly m characters
and represent the field. 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 first field line where n = m = 0 represents the end of input and should
not be processed.
The Output
For each field, you must print the following message in a line alone:
Field #x:
Where x stands for the number of the field (starting from 1). The next n lines should contain the field with the "." characters replaced by the number of adjacent mines to that square. There must be an empty line between field outputs.
Sample Input
4 4 *... .... .*.. .... 3 5 **... ..... .*... 0 0
Sample Output
Field #1: *100 2210 1*10 1110 Field #2: **100 33200 1*100
位向量解决
#include <stdio.h>
#include <string.h>
#define max 100
char mine[max][max];
int main()
{
int m, n, i, j, k, count = 1;
int px[8] = {-1,-1,-1, 0, 0, 1, 1, 1};
int py[8] = {-1, 0, 1,-1, 1,-1, 0, 1};
while (scanf("%d %d", &n, &m) && (m || n))
{
memset(mine, '.', sizeof(mine));
for (i = 0; i < n; i++)
for (j = 0; j < m; j++)
scanf(" %c", &mine[i][j]);
for (i = 0; i < n; i++)
for (j = 0; j < m; j++)
{
int tmp = 0;
if (mine[i][j] == '.')
for (k = 0; k < 8; k++)
{
if (mine[i+px[k]][j+py[k]] == '*')
tmp++;
mine[i][j] = tmp + '0';
}
}
if (count != 1)
printf("\n");
printf("Field #%d:\n", count++);
for (i = 0; i < n; i++)
{
for (j = 0; j < m; j++)
printf("%c", mine[i][j]);
printf("\n");
}
}
return 0;
}