蓝桥每周练二扫雷问题（坐标的移动）

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本文详细解析了扫雷游戏的算法实现，通过定义移动数组和遍历字符数组的方法，计算每个安全方格周围八方位的地雷数量，最终输出二维数组表示的地雷分布情况。

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Minesweeper （扫雷）

[问题描述]

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 M × N 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 4 × 4 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.

[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 ≤ 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.

【题目大意】扫雷游戏，输入一块二维的字符区域，用*表示地雷，然后输出int类型的二维数组，num[i][j]表示与它邻近的八个方位(东，南，西北，东北…)有的地雷数

[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

【思路】定义一个移动数组int dx[3]={-1,0,1}; int dy[3]={0,1,-1};，遍历字符数组，遇到‘*’，则与它邻近的八个点 num[i+dx[a]][j+dy[b]]++
【代码】

#include <iostream>
#include<string.h>
using namespace std;
int dx[3]={-1,0,1};
int dy[3]={0,1,-1};
int n,m;
const int INF=-999;
int main()
{

   cin>>n>>m;
   char mine[n][m];
   int  num[n][m];
    memset(num, 0, sizeof(num));//扫雷数初始化为0
/***构造字符数组***/
   for(int i=0;i<n;i++)
   {
       for(int j=0;j<m;j++)
       {
           cin>>mine[i][j];
       }
   }


    for(int i=0;i<n;i++)
   {
       for(int j=0;j<m;j++)
       {
           if(mine[i][j]=='*')
           {
               for(int a=0;a<3;a++)
               {
                for(int b=0;b<3;b++)
                {
					if(((i+dx[a])>=0)&&((i+dx[a])<n)&&((j+dy[b])>=0)&&((i+dy[b])<m)&&(dy[a]!=0||dx[a]!=0))//没有超过边界且不是自身
                        num[i+dx[a]][j+dy[b]]++;
                }
               num[i][j]=INF;//自身无穷小，用于标记，等会儿输出“*”

               }


           }
       }
   }

    for(int i=0;i<n;i++)//按格式输出
   {
       for(int j=0;j<m;j++)
       {
           if(num[i][j]<0)
                cout<<'*';
         else
            cout<<num[i][j];
       }
       cout<<endl;
   }

    return 0;
}