描述

I think most of you are using system named of xp or vista or win7.And these system is consist of a famous game what is mine sweeping.You must have played it before.If you not,just look the game rules followed.

There
are N*N grids on the map which contains some mines , and if you touch
that ,you lose the game.If a position not containing a mine is touched,
an integer K (0 < =K <= 8) appears indicating that there are K
mines in the eight adjacent positions. If K = 0, the eight adjacent
positions will be touched automatically, new numbers will appear and
this process is repeated until no new number is 0. Your task is to mark
the mines' positions without touching them.

Now, given the distribution of the mines, output the numbers appearing after the player's first touch.

输入

The
first line of each case is two numbers N (1 <= N <= 100) .Then
there will be a map contain N*N grids.The map is just contain O and
X.'X' stands for a mine, 'O' stand for it is safe with nothing. You can
assume there is at most one mine in one position. The last line of each
case is two numbers X and Y(0<=X<N,0<=Y<N, indicating the
position of the player's first touch.

输出

If the player touches the mine, just output "it is a beiju!".

If
the player doesn't touch the mine, output the numbers appearing after
the touch. If a position is touched by the player or by the computer
automatically, output the number. If a position is not touched, output a
dot '.'.

Output a blank line after each test case.

样例输入

5
OOOOO
OXXXO
OOOOO
OXXXO
OOOOO
1 1
5
OOOOO
OXXXO
OOOOO
OXXXO
OOOOO
0 0

样例输出

it is a beiju!

1....
.....
.....
.....
.....

题目来源

HDOJ

扫雷游戏~

#include <stdio.h>
#include <string.h>
#define MAXN 150
int N;
int flag[MAXN][MAXN];
char map[MAXN][MAXN];
char out[MAXN][MAXN];
int judge(int x, int y){
int flag=0;
for(int i=x-1; i<=x+1; i++){
for(int j=y-1; j<=y+1; j++){
if(1<=i && i<=N && 1<=j && j<=N && !(i==x&&j==y)){
if(map[i][j]=='X')flag++;
}
}
}
return flag;
}
void dfs(int x, int y){
int sum=judge(x,y);
if(sum==0){
out[x][y]='0';
for(int i=x-1; i<=x+1; i++){
for(int j=y-1; j<=y+1; j++){
if(1<=i && i<=N && 1<=j && j<=N && !(i==x&&j==y)){
if(!flag[i][j])
dfs(i,j);
}
flag[i][j]=1;
}
}
}else{
out[x][y]=sum+'0';
}
}
int main()
{
while( scanf("%d",&N)!=EOF ){
for(int i=1; i<=N; i++){
getchar();
for(int j=1; j<=N; j++){
scanf("%c",&map[i][j]);
}
}
int x,y;
scanf("%d %d",&x,&y);
x++;
y++;
if(map[x][y]=='X'){
puts("it is a beiju!\n");
continue;
}
memset(flag,0,sizeof(flag));
memset(out,'.',sizeof(out));
dfs(x,y);
for(int i=1; i<=N; i++){
for(int j=1; j<=N; j++){
printf("%c",out[i][j]);
}
printf("\n");
}
printf("\n");
}
return 0;
}

TOJ 3184 Mine sweeping的更多相关文章

  1. 8659 Mine Sweeping

    时间限制:500MS  内存限制:65535K提交次数:37 通过次数:15 题型: 编程题   语言: G++;GCC Description The opening ceremony of the ...

  2. 【HDOJ】3316 Mine sweeping

    简单BFS. #include <iostream> #include <cstdio> #include <cstring> #include <cstdl ...

  3. Mine Number(搜索,暴力) ACM省赛第三届 G

    Mine Number Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Every one once played the gam ...

  4. [2012山东省第三届ACM大学生程序设计竞赛]——Mine Number

    Mine Number 题目:http://acm.sdut.edu.cn/sdutoj/problem.php? action=showproblem&problemid=2410 Time ...

  5. C language 模拟 win的经典游戏——扫雷

    让我们在terminal下愉快的...扫雷 昨天跟奇葩霖聊起"雷区"这个敏感词汇,然后非常荣幸的... 应该轰炸不到我.. . 后来百无聊赖的去玩了把扫雷.然后发现我之前都是乱扫的 ...

  6. SVN版本冲突,导致出现Files 的值“ < < < < < < < .mine”无效

    只要根据错误提示,找到相应文件夹下的\obj\Debug文件夹下的 相应名字.csproj.FileListAbsolute.txt, 打开并删除含有'<<<<<< ...

  7. Files 的值“<<<<<<< .mine”无效。路径中具有非法字符

    解决冲突,告诉SVN这个问题已解决(Resolved). 一般更简单些:在你的工程OBJ/DEBUG目录下,找到 工程名.csproj.FileListAbsolute.txt的文件打开并删除含有'& ...

  8. SVN Files 的值“ < < < < < < < .mine”无效。路径中具有非法字符。

    错误 1 Files 的值“ < < < < < < < .mine”无效.路径中具有非法字符.     今天使用SVN进行更新的时候,出现了如上问题,想起卓 ...

  9. TOJ 2776 CD Making

    TOJ 2776题目链接http://acm.tju.edu.cn/toj/showp2776.html 这题其实就是考虑的周全性...  贡献了好几次WA, 后来想了半天才知道哪里有遗漏.最大的问题 ...

随机推荐

  1. jQuery事件(持续更新中)

    方法 描述 bind() 向匹配元素附加一个或更多事件处理器 blur() 触发.或将函数绑定到指定元素的 blur 事件 change() 触发.或将函数绑定到指定元素的 change 事件 cli ...

  2. Android中如何下载文件并显示下载进度

    原文地址:http://jcodecraeer.com/a/anzhuokaifa/androidkaifa/2014/1125/2057.html 这里主要讨论三种方式:AsyncTask.Serv ...

  3. IIS 6.0 发布网站使用教程

    原文地址:http://wenku.baidu.com/view/95d8b49851e79b89680226aa.html

  4. WinForm中的重绘 - 文本的重绘

    两种方式 TextRenderer.DrawText 注意:默认在每次绘制的文本左右有padding,即使参数中设置了TextFormatFlags.NoPadding也是一样,因此在分段绘制文本时( ...

  5. SpringMVC+Hibernate 项目开发之二 (STS整合Maven)

    为什么用STS不用Eclipse,主要是Eclipse集成Maven把我整疯了,最后估计原因除在网速上了. 其实用了STS以后发现还真比Eclipse好用点. STS本身集成有Maven的,但是默认的 ...

  6. linux获取域名地址

    dig live-195887137.cn-north-1.elb.amazonaws.com.cn +short

  7. fiddler 代理调试本地手机页面

    https://www.cnblogs.com/zichi/p/4944581.html

  8. loj#6041. 「雅礼集训 2017 Day7」事情的相似度(后缀自动机+启发式合并)

    题面 传送门 题解 为什么成天有人想搞些大新闻 这里写的是\(yyb\)巨巨说的启发式合并的做法(虽然\(LCT\)的做法不知道比它快到哪里去了--) 建出\(SAM\),那么两个前缀的最长公共后缀就 ...

  9. python爬虫的一些小小问题、python动态正则表达式

    1.首先urllib不能用了,需要引入的是urllib2,正则re. #coding=utf-8 # import urllib import urllib2 import re def getHtm ...

  10. Android---16进制与字节数组

    16进制字符串与字节数组进行转换 package string; import java.util.Arrays; /** * byte[]与16进制字符串相互转换 * * @date:2017年4月 ...