UVa10189 Minesweeper

 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 tofind where are all the mines within a MxN field. To help you, the game shows a numberin 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:

*10022101*101110

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 containstwo integers n and m (0 < n,m <= 100) which stands for the number of lines and columns of the fieldrespectively. The next n lines contains exactly m characters and represent the field. Each safesquare is represented by an "." character (without the quotes) and each mine square is representedby 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 shouldcontain the field with the "." characters replaced by the number of adjacent minesto that square. There must be an empty line between field outputs.

Sample Input

 

4 4*........*......3 5**.........*...0 0

 

Sample Output

 

Field #1:*10022101*101110Field #2:**100332001*100

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

UVa上A掉的第二个题。。。

思路:思路很简单就是从头到尾一次遍历一遍读进来的数据，遇到'*'就在一个对应的表示结果的二维数组res中向周围8个方向上依次做相应的地雷数统计操作。最后在遍历一遍二维数组res把结果按要求输出就可以了。

Code:

#include<stdio.h>#include<string.h>int res[102][102];char str[102][102];int main(){ int cases=0; int m,n; int i,j; while(scanf("%d%d",&m,&n)) { if(m==0&&n==0) { break; } memset(res,0,sizeof(res)); cases++; if(cases!=1) { printf("/n"); } for(i=1;i<=m;i++) { scanf("%s",str[i]); } for(i=1;i<=m;i++) for(j=0;j<n;j++) { if(str[i][j]=='*') { res[i][j+1]=-99999; res[i-1][j+1]+=1; res[i+1][j+1]+=1; res[i][j+2]+=1; res[i][j]+=1; res[i-1][j]+=1; res[i-1][j+2]+=1; res[i+1][j]+=1; res[i+1][j+2]+=1; } } printf("Field #%d:/n",cases); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("%c",res[i][j]>=0?'0'+res[i][j]:'*'); } printf("/n"); } } return 0;}