番3 Minesweeper (使用int型数组代替char型数组 ) （C语言代码）

解题思路:   
                使用了轰炸法,参考了 16计三梁海栩     的代码,即为发现一个雷就将周围的8个区域的值加一.

                统一使用int 型的二维数组代表*和.     解决了字符型数据和数字难以保存在一起的问题

                由于每个格子周围最多有8个雷,故格子里的数值不会超过8,可以用一个大于8的数来代表雷,这样就

                可以在爆雷加1的时候不像char型数据那样会影响到*的保存,在遍历打印的时候只需要判断一下就

                可以知道是不是雷了.

                如果有兴趣的话可以指点一下我最下方注释的代码,那是我起初打出来的代码,能过测试样例,自己写的样例也能过,但是就是过不了OJ系统.
参考代码:

/* 题目描述
Minesweeper 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 find where all the mines are located within a M x N field. 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 x 4 field on the left contains two mines, each represented by a ``*'' character. If we represent the same field by the hint numbers described above, we end up with the field on the right: *... .... .*.. .... *100 2210 1*10 1110
输入
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$ \le$100) which stand for the number of lines and columns of the field, respectively. Each of the next n lines contains exactly m characters, representing the field. Safe squares are denoted by ``.'' and mine squares by ``*,'' both without the quotes. The first field line where n = m = 0 represents the end of input and should not be processed.
输出
For each field, print the message Field #x: on a line alone, 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 mines adjacent to that square. There must be an empty line between field outputs.
样例输入
4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0
样例输出
Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100 */




#include <stdio.h>
#include <string.h> 

int main (){
    int n,m;
    int count=0;        //给输出时候的filed编号用
    while (1){
        int i,j;
        int a[102][102];            //定义镶边的二维数组,包围着的边用用来防止给*周围的格子计数时候越界
        for (i=0;i<102;i++){        //使用for循环给二维数组赋初始值0;
            for (j=0;j<102;j++){
                a[i][j]=0;
            }
        }
    scanf("%d%d",&n,&m);        //输入n*m的矩阵
        getchar();              //吞掉回车
        if (n==0&&m==0){
            break;              //当输入m=0,n=0时候结束
        }
        count++;
        char str[102];          //定义一个临时的字符型数组用来给矩阵的每一行赋值
        
        for (i=1;i<=n;i++){
            scanf ("%s",&str[1]);       //从键盘获得一串字符,并且自动忽略回车,从1开始为了与下面a[i][j]对应
            for (j=1;j<=m;j++){
                if (str[j]=='*'){
                    a[i][j]=99;         //如果将要读入矩阵的字符是*,赋值为一个不可能的数代表为*(因为当一个空格周围最多为8个地雷,故最大为8,大于8即为不可能的数),输出的时候判断用
                    int k,l;                   //定义k,l继承i,j,避免改变i,j的值影响到后面
                    for (k=i-1;k<i+2;k++){      //给这个地雷的周边8个格子都加1;
                        for (l=j-1;l<j+2;l++){
                            if (a[k][l]<99){        //判断一下周围8个格子是否为*,给不为*的加1
                                a[k][l]++;
                            }
                        }
                    }
                }
            }
        }
    printf ("Field #%d:\n",count);          //打印表头
    for (i=1;i<=n;i++){
        for (j=1;j<=m;j++){
            if (a[i][j]<99){        //打印出来不为*的值
                printf ("%d",a[i][j]);
            }
            else {
                printf("*");        //打印出来*,(大于99)
            }
        }
        printf ("\n");
    }
    printf ("\n");  
    }
    return 0;
}



//用字符型二维数组做的,给8个边界值与中间值分别处理(过于麻烦),但是做出来的样例过了,不过过不了oj系统
//不知道为什么,自测数据也能通过,就是过不了oj
/* #include <stdio.h>
char a[100][100];            //定义二维数组
int change(int n,int m){
int i,j;
for (i=0;i<n;i++){
        for (j=0;j<m;j++){
            if (a[i][j]=='*'){      //对于*的格子,给边界格子分别+1
                if (i==0&&j==0){            分别处理八个边界和一个中间的分类
                    if (a[i][j+1]!='*'){
                        a[i][j+1]+=1;
                    }
                    if (a[i+1][j]!='*'){
                        a[i+1][j]+=1;
                    }
                    if (a[i+1][j+1]!='*'){
                        a[i+1][j+1]+=1;
                    }                                       
                }
                if (i==0&&j<m-1&&j!=0){
                    if (a[i][j-1]!='*'){
                        a[i][j-1]+=1;
                    }
                    if (a[i+1][j-1]!='*'){
                        a[i+1][j-1]+=1;
                    }
                    if (a[i+1][j]!='*'){
                        a[i+1][j]+=1;
                    }
                    if (a[i+1][j+1]!='*'){
                        a[i+1][j+1]+=1;
                    }
                    if (a[i][j+1]!='*'){
                        a[i][j+1]+=1;
                    }
                }
                if (i==0&&j==m-1){
                    if (a[i][j-1]!='*'){
                        a[i][j-1]+=1;
                    }
                    if (a[i+1][j-1]!='*'){
                        a[i+1][j-1]+=1;
                    }
                    if (a[i+1][j]!='*'){
                        a[i+1][j]+=1;
                    }                                       
                }
                if (i<n-1&&j==0&&i!=0){
                    if (a[i-1][j]!='*'){
                        a[i-1][j]+=1;
                    }
                    if (a[i-1][j+1]!='*'){
                        a[i-1][j+1]+=1;
                    }
                    if (a[i][j+1]!='*'){
                        a[i][j+1]+=1;
                    }
                    if (a[i+1][j+1]!='*'){
                        a[i+1][j+1]+=1;
                    }
                    if (a[i+1][j]!='*'){
                        a[i+1][j]+=1;
                    }                                       
                }
                if (i==n-1&&j==0){
                    if (a[i-1][j]!='*'){
                        a[i-1][j]+=1;
                    }
                    if (a[i-1][j+1]!='*'){
                        a[i-1][j+1]+=1;
                    }
                    if (a[i][j+1]!='*'){
                        a[i][j+1]+=1;
                    }                                       
                }
                if (i==n-1&&j<m-1&&j!=0){
                    if (a[i][j-1]!='*'){
                        a[i][j-1]+=1;
                    }
                    if (a[i-1][j-1]!='*'){
                        a[i-1][j-1]+=1;
                    }
                    if (a[i-1][j]!='*'){
                        a[i-1][j]+=1;
                    }
                    if (a[i-1][j+1]!='*'){
                        a[i-1][j+1]+=1;
                    }
                    if (a[i][j+1]!='*'){
                        a[i][j+1]+=1;
                    }                                       
                }
                if (i==n-1&&j==m-1){
                    if (a[i][j-1]!='*'){
                        a[i][j-1]+=1;
                    }
                    if (a[i-1][j-1]!='*'){
                        a[i-1][j-1]+=1;
                    }
                    if (a[i-1][j]!='*'){
                        a[i-1][j]+=1;
                    }
                                                            
                }
                if (i<n-1&&j==m-1&&i!=0){
                    if (a[i-1][j]!='*'){
                        a[i-1][j]+=1;
                    }
                    if (a[i-1][j-1]!='*'){
                        a[i-1][j-1]+=1;
                    }
                    if (a[i][j-1]!='*'){
                        a[i][j-1]+=1;
                    }
                    if (a[i+1][j-1]!='*'){
                        a[i+1][j-1]+=1;
                    }
                    if (a[i+1][j]!='*'){
                        a[i+1][j]+=1;
                    }
                                                                
                }
                     if (i>0&&i<n-1&&j>0&&j<m-1) {      //对于中间的采取了继承的做法,参考上面那个成功的代码
                        int k,l;
                        for (k=i-1;k<i+2;k++){
                            for (l=j-1;l<j+2;l++){
                                if (a[k][l]!='*'){
                                    a[k][l]++;
                                }
                            }
                        }
                    }
            }
        }
    }
}
int main ()
{
int n;
int m;
int count;
int i,j;while ( scanf( "%d %d", &n, &m ) && !(n == 0 && m == 0) ){
    count++;
    getchar();
    for (i=0;i<n;i++){
        for (j=0;j<m;j++){
            scanf ("%c",&a[i][j]);      //读入字符并且改变字符
            if (a[i][j]=='.'){
                a[i][j]='0';
            }
        }
        getchar();
    }
    change(n,m);            //调用处理格子的函数
    printf ("Field #%d:\n",count);      //打印表头
    for (i=0;i<n;i++){
        for (j=0;j<m;j++){
            printf ("%c",a[i][j]);
        }
        printf ("\n");
    }
    printf ("\n");  
}

return 0;   
} */