UVA 10189   C - MINES p

C - MINES p 

Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu  

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Description

 Problem B: Minesweeper  

The Problem

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 MxN 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 4x4 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. 

The 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 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. 

The Output

For each field, you must print the following message in a line alone: 

Field #x: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 adjacent mines 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

--------------------------------------------------------------------------------

? 2001 Universidade do Brasil (UFRJ). Internal Contest Warmup 2001. 

Input

Output

Sample Input

Sample Output

Hint

[Submit]   [Go Back]   [Status] 

这题其实挺简单的，但是就是一个空行的位置让我纠结了半天也没过，注意啊注意

代码：

#include<stdio.h>

#include<string.h>

int main()

{

char map[202][202];

int num[202][202],k,s,kx,ky,i,j,m,n,sum=0,flag[8][2]={-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};

scanf("%d%d",&m,&n);

if(m==0&&n==0)return 0;

while(1)

{

sum++;

memset(num,0,sizeof(num));

for(i=0;i<m;i++)

scanf("%s",map[i]);

printf("Field #%d:\n",sum);

for(i=0;i<m;i++)

{

for(j=0;j<n;j++)

if(map[i][j]=='*')

printf("*");

else{

s=0;

for(k=0;k<8;k++)

{

kx=i+flag[k][0];

ky=j+flag[k][1];

if(kx>=0&&kx<m&&ky>=0&&ky<n&&map[kx][ky]=='*')

s++;

}

printf("%d",s);

}

printf("\n");

}

scanf("%d%d",&m,&n);

if(m==0&&n==0)

break;

else printf("\n");

}

return 0;

}