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PC/UVA 110102/10189 |
水题。看样例输入输出即可明白题意。我的做法是用一个hash直接处理就成
//author: CHC
//First Edit Time: 2014-01-10 23:39
//Last Edit Time: 2014-01-10 23:39
//Filename:1.cpp
#include <iostream>
#include <cstdio>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
int n,m;
char map[120][120];
int main()
{
int cas=0;
int flag=0;
while(~scanf("%d%d",&n,&m)&&(n||m))
{
if(flag)puts("");
memset(map,'0',sizeof(map));
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
char x;
scanf(" %c",&x);
if(x=='*')
{
map[i][j]='*';
for(int ii=i-1;ii<=i+1;ii++)
for(int jj=j-1;jj<=j+1;jj++)
if(map[ii][jj]!='*')++map[ii][jj];
}
}
printf("Field #%d:\n",++cas);
for(int i=1;i<=n;i++,puts(""))
for(int j=1;j<=m;j++)
putchar(map[i][j]);
flag=1;
}
return 0;
}
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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