C语言扫雷程序

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 4void 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]);elseprintf("%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);}