采用的算法比较死板，太多if...else语句结合，希望有大牛点评一下下哈！
朋友说可以把雷区矩阵一次性开辟大一点，然后把外围全填零，就可以避免判断端点、边界，只要全部判断八方向就好了。
/**
i:行
j:列
四连通[ 上(i-1,j)、 下(i+1,j)、 左(i,j-1)、 右(i,j+1)]
八连通[ 左上(i-1,j-1)、左下(i+1,j-1)、右上(i-1,j+1)、右下(i+1,j+1) ]
**/
#include <stdio.h>
#define M 4
#define N 4
void output(char op[M][N])
{
for(int i=0;i<M;i++)
{
for(int j=0;j<N;j++)
{
if(op[i][j] == '*')
printf("%c ",op[i][j]);
else
printf("%d ",op[i][j]-'0');
}
printf("\n");
}
}
void saolei(char b[M][N])
{
int i,j;
for(i=0;i<M;i++)
{
for(j=0;j<N;j++)
{
if(b[i][j] != '*')
{
b[i][j]='0';
if(i==0 && j==0)//左上顶点
{
if(b[i+1][j]=='*') b[i][j]++;
if(b[i][j+1]=='*') b[i][j]++;
if(b[i+1][j+1]=='*') b[i][j]++;
continue;
}
else if(i==M-1 && j==0)//左下顶点
{
if(b[i-1][j] == '*') b[i][j]++;
if(b[i][j+1] == '*') b[i][j]++;
if(b[i-1][j+1] == '*') b[i][j]++;
continue;
}
else if(i==0 && j==N-1)//右上顶点
{
if(b[i][j-1] == '*') b[i][j]++;
if(b[i+1][j] == '*') b[i][j]++;
if(b[i+1][j-1] == '*') b[i][j]++;
continue;
}
else if(i==M-1 && j==N-1)//右下顶点
{
if(b[i][j-1] == '*') b[i][j]++;
if(b[i-1][j] == '*') b[i][j]++;
if(b[i-1][j-1] == '*') b[i][j]++;
continue;
}
else if(j==0)//左边界
{
if(b[i-1][j] == '*') b[i][j]++;
if(b[i+1][j] == '*') b[i][j]++;
if(b[i][j+1] == '*') b[i][j]++;
if(b[i-1][j+1] == '*') b[i][j]++;
if(b[i+1][j+1] == '*') b[i][j]++;
continue;
}
else if(j==N-1)//右边界
{
if(b[i-1][j] == '*') b[i][j]++;
if(b[i+1][j] == '*') b[i][j]++;
if(b[i][j-1] == '*') b[i][j]++;
if(b[i-1][j-1] == '*') b[i][j]++;
if(b[i+1][j-1] == '*') b[i][j]++;
continue;
}
else if(i==0)//上边界
{
if(b[i][j-1] == '*') b[i][j]++;
if(b[i][j+1] == '*') b[i][j]++;
if(b[i+1][j] == '*') b[i][j]++;
if(b[i+1][j+1] == '*') b[i][j]++;
if(b[i+1][j-1] == '*') b[i][j]++;
continue;
}
else if(i==M-1)//下边界
{
if(b[i][j-1] == '*') b[i][j]++;
if(b[i][j+1] == '*') b[i][j]++;
if(b[i-1][j] == '*') b[i][j]++;
if(b[i-1][j+1] == '*') b[i][j]++;
if(b[i-1][j-1] == '*') b[i][j]++;
continue;
}
else
{
if(b[i-1][j] == '*') b[i][j]++;
if(b[i+1][j] == '*') b[i][j]++;
if(b[i][j-1] == '*') b[i][j]++;
if(b[i][j+1] == '*') b[i][j]++;
if(b[i+1][j+1] == '*') b[i][j]++;
if(b[i-1][j-1] == '*') b[i][j]++;
if(b[i+1][j-1] == '*') b[i][j]++;
if(b[i-1][j+1] == '*') b[i][j]++;
continue;
}
}
}
}
}
void main()
{
char a[M][N]={{'*','.','.','.'}, {'.','.','.','.'}, {'.','*','.','.'}, {'.','.','.','.'}};
printf("初始为4*4的雷区矩阵(*代表地雷，.代表无地雷):\n");
for(int i=0;i<M;i++)
{
for(int j=0;j<N;j++)
{
printf("%c ",a[i][j]);
}
printf("\n");
}
saolei(a);
printf("经计算机扫描后的结果:\n");
output(a);
}