Minesweeper

题意：

Have you ever played Minesweeper? This cute little game comes with a certain operating system
whose name we can't remember. The goal of the game is to nd where all the mines are located within
a M  N eld.
The game shows a number in a square which tells you how many mines there are adjacent to that
square. Each square has at most eight adjacent squares. The 4 4 eld on the left contains two mines,
each represented by a \*" character. If we represent the same eld by the hint numbers described
above, we end up with the eld on the right:
*...
....
.*..
....
*100
2210
1*10
1110
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 stand for the number of lines and columns of the eld,
respectively. Each of the next n lines contains exactly m characters, representing the eld.
Safe squares are denoted by \." and mine squares by \*," both 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, print the message Field #x: on a line alone, 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 mines adjacent 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

解题思路：

建立一个数组，以某一个点作为远点，然后旁边八个点作为左边，形成一个【8】【2】的二维数组，然后依次找出每个点的值

代码：

#include <stdio.h>
 #define MAXSIZE 100
  int  range_checking(int x, int y, int line, int row)  
	{  
	    return ((0 <= x && x < line) && (0 <= y && y < row));  
	} 
    
 void display(char matrix[][MAXSIZE], int line, int row)  
	{  
        int i,j,k,b;           
	    int bounds[8][2] =  
	        { {-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0},  
	        {1, 1} };  
	  
	    for ( i = 0; i < line; i++)  
	    {  
	        for ( j = 0; j < row; j++)  
	        {  
	            // 如果该位置为地雷则原样输出地雷的符号。  
	            if (matrix[i][j] == '*')  
	                printf( "*");  
	            else  
	            {  
	                // 统计该点周围 8 点的地雷总数。  
	                int mines = 0;  
	                for ( k = 0; k < 8; k++)  
	                {  
	                    int m = i + bounds[k][0];  
	                    int n = j + bounds[k][1];  
	                    b = range_checking(m, n, line, row);
	                    if (b && matrix[m][n] == '*')  
	                        mines++;  
	                }  
	  
	                printf("%d",mines);  
	            }  
	        }  
	  
	   printf("\n");  
	    }  
	} 
    
    int main()  
	{  
	    char matrix[MAXSIZE][MAXSIZE];  
	    int line,i,j,row, field = 0;  
	    scanf("%d %d",&line,&row);
	    while ( line>0&& line <= 100 && row > 0 && row <= 100)  
	    {  
	        //memset(matrix, 0, sizeof(matrix));  
	  
	        for ( i = 0; i < line; i++)  
	            for ( j = 0; j < row; j++)  
	               //printf("%c", matrix[i][j]);
                     scanf("%c", &matrix[i][j]);
  
	  
	     
	        field++;  
	  
	        printf("Field #%d:\n",field);
	  
	        display(matrix, line, row);
                
                 scanf("%d %d",&line,&row);
      }        system("pause"); }    

但是出现了错误，至今没能找出。