烟大 Contest1024 - 《挑战编程》第一章：入门 Problem B: Minesweeper（模拟扫雷）

Problem B: Minesweeper

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 29  Solved: 7
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Description

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

Input

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 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.

Output

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.

Sample Input

4 4

*...

....

.*..

....

3 5

**...

.....

.*...

0 0

Sample Output

Field #1:

*100

2210

1*10

1110

 

 

Field #2:

**100

33200

1*100

HINT

---

　　模拟题。模拟的是扫雷这个小游戏。输入雷的分布，输出每一个格子四周的雷数矩阵。也就是模拟扫雷里数字的产生过程。

　　一开始把计算那一步想的复杂了，由于没有初始化，我只好把检测的情况分为边缘、内部，边缘又分为四角和边，这样有三种情况，每一种情况的检测区域又不一样，我要每一种都分别定义。后来发现没必要分类，直接检测周围8个格子的区域就行，只不过这样就需要提前初始化好以及让数组留出边缘。

My code：

 #include <iostream>

 using namespace std;

 int main()
 {
     char a[][]={};
     int n,m,count=;
     while(cin>>n>>m){
         if(n== && m==)
             break;
         for(int i=;i<=n;i++)   //input
             for(int j=;j<=m;j++)
                 cin>>a[i][j];
         for(int i=;i<=n;i++)   //计算
             for(int j=;j<=m;j++){
                 if(a[i][j]=='*')
                     continue;

                 a[i][j]='';
                 if(a[i-][j]=='*')
                     ++a[i][j];
                 if(a[i+][j]=='*')
                     ++a[i][j];
                 if(a[i][j+]=='*')
                     ++a[i][j];
                 if(a[i][j-]=='*')
                     ++a[i][j];

                 if(a[i-][j-]=='*')
                     ++a[i][j];
                 if(a[i-][j+]=='*')
                     ++a[i][j];
                 if(a[i+][j-]=='*')
                     ++a[i][j];
                 if(a[i+][j+]=='*')
                     ++a[i][j];
             }

         if(count!=)
             cout<<endl;
         cout<<"Field #"<<count++<<':'<<endl;
         for(int i=;i<=n;i++){  //输出
             for(int j=;j<=m;j++){
                 cout<<a[i][j];
             }
             cout<<endl;
         }
     }
     return ;
 }

 /**************************************************************
     Problem: 1099
     User: freecode
     Language: C++
     Result: Accepted
     Time:0 ms
     Memory:1268 kb
 ****************************************************************/

Freecode : www.cnblogs.com/yym2013