烟大 Contest1024

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：

 1 #include <iostream>
 2 
 3 using namespace std;
 4 
 5 int main()
 6 {
 7     char a[102][102]={0};
 8     int n,m,count=1;
 9     while(cin>>n>>m){
10         if(n==0 && m==0)
11             break;
12         for(int i=1;i<=n;i++)   //input
13             for(int j=1;j<=m;j++)
14                 cin>>a[i][j];
15         for(int i=1;i<=n;i++)   //计算
16             for(int j=1;j<=m;j++){
17                 if(a[i][j]=='*')
18                     continue;
19 
20                 a[i][j]='0';
21                 if(a[i-1][j]=='*')
22                     ++a[i][j];
23                 if(a[i+1][j]=='*')
24                     ++a[i][j];
25                 if(a[i][j+1]=='*')
26                     ++a[i][j];
27                 if(a[i][j-1]=='*')
28                     ++a[i][j];
29 
30                 if(a[i-1][j-1]=='*')
31                     ++a[i][j];
32                 if(a[i-1][j+1]=='*')
33                     ++a[i][j];
34                 if(a[i+1][j-1]=='*')
35                     ++a[i][j];
36                 if(a[i+1][j+1]=='*')
37                     ++a[i][j];
38             }
39 
40         if(count!=1)
41             cout<<endl;
42         cout<<"Field #"<<count++<<':'<<endl;
43         for(int i=1;i<=n;i++){  //输出
44             for(int j=1;j<=m;j++){
45                 cout<<a[i][j];
46             }
47             cout<<endl;
48         }
49     }
50     return 0;
51 }
52 
53 /**************************************************************
54     Problem: 1099
55     User: freecode
56     Language: C++
57     Result: Accepted
58     Time:0 ms
59     Memory:1268 kb
60 ****************************************************************/

Freecode : www.cnblogs.com/yym2013