蓝桥杯提高题 Minesweeper

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: ∗........∗......∗10022101∗101110*... .... .*.. .... *100 2210 1*10 1110∗........∗......∗10022101∗101110

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)( 0 < n, m\leq100)(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
1110Field #2:
**100
33200
1*100

简单BFS，注意输出格式即可：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
char s[100][100];
int num[100][100],n,m;
int step[8][2]={0,1, 0,-1, 1,0 ,-1,0, 1,1, -1,1, 1,-1, -1,-1};void bfs(int x,int y){int xx,yy;for(int i=0;i<8;i++){xx=x+step[i][0];yy=y+step[i][1];if(xx>=0 && xx<4 && yy>=0 && yy<4 && s[xx][yy]!='*') num[xx][yy]++;}
}int main() {int id=0;while(cin>>n>>m){id++;if(n==0 && m==0) break;for(int i=0;i<n;i++){for(int j=0;j<m;j++){cin>>s[i][j];}}fill(num[0],num[0]+100*100,0);for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(s[i][j]=='*') bfs(i,j);}}printf("Field #%d:\n",id);for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(s[i][j]=='*') cout<<"*";else cout<<num[i][j];}puts("");}puts("");}
}