题目描述
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
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 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.
输出
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.
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
参考代码
#include <stdio.h>
int f(int a)
{
if(a==-1)return 1; else return 0;
}
int main()
{
/* freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);*/
int a,b,count=1,first=1;
while(scanf("%d%d",&a,&b)==2 && (a!=0 || b!=0))
{
if(first)first=0; else printf("nn");
int i,j;
char s[110][110];
int h[110][110]= {0};
for(i=0; i<=a; i++)gets(s[i]);
for(i=1; i<=a; i++)
{
for(j=1; j<=b; j++)
{
if(s[i][j-1]=='*')h[i][j]=-1;
else h[i][j]=0;
}
}
for(i=1; i<=a; i++)
{
for(j=1; j<=b; j++)
{
if(h[i][j]!=-1)
{
h[i][j]=f(h[i-1][j-1])+f(h[i-1][j])+f(h[i-1][j+1])+f(h[i][j-1])+f(h[i][j+1])+f(h[i+1][j-1])+f(h[i+1][j])+f(h[i+1][j+1]);
}
}
}
printf("Field #%d:n",count++);
for(i=1; i<=a; i++)
{
for(j=1; j<=b; j++)
{
if(h[i][j]==-1)printf("*");
else printf("%d",h[i][j]);
}
if(i!=a)printf("n");
}
}
printf("n");
return 0;
}
解析
暂无
