Codeforces Round #290 (Div. 2) B. Fox And Two Dots dfs
B. Fox And Two Dots
题目连接:
http://codeforces.com/contest/510/problem/B
Description
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
These k dots are different: if i ≠ j then di is different from dj.
k is at least 4.
All dots belong to the same color.
For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists a cycle, and "No" otherwise.
Sample Input
3 4
AAAA
ABCA
AAAA
Sample Output
Yes
Hint
题意
给你一个n*m的网格
然后问你是否有只含有一种元素的环
题解:
dfs就好了
dfs的时候,记录一下fa,然后一直跑下去,跑到曾经vis过的地方,就说明遇到了环
代码
#include<bits/stdc++.h>
using namespace std;
char mp[55][55];
int vis[55][55];
int flag = 0;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int n,m;
void dfs(int x,int y,char c,int fax,int fay)
{
vis[x][y]=1;
if(flag)return;
for(int i=0;i<4;i++)
{
int xx = x+dx[i];
int yy = y+dy[i];
if(xx==fax&&yy==fay)continue;
if(xx<0||xx>=n)continue;
if(yy<0||yy>=m)continue;
if(mp[xx][yy]!=c)continue;
if(vis[xx][yy]){
flag=1;
return;
}
dfs(xx,yy,c,x,y);
}
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%s",&mp[i]);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(!vis[i][j])
dfs(i,j,mp[i][j],i,j);
}
if(flag)printf("Yes");
else printf("No\n");
}
Codeforces Round #290 (Div. 2) B. Fox And Two Dots dfs的更多相关文章
- Codeforces Round #290 (Div. 2) B. Fox And Two Dots(DFS)
http://codeforces.com/problemset/problem/510/B #include "cstdio" #include "cstring&qu ...
- DFS Codeforces Round #290 (Div. 2) B. Fox And Two Dots
题目传送门 /* DFS:每个点四处寻找,判断是否与前面的颜色相同,当走到已走过的表示成一个环 */ #include <cstdio> #include <iostream> ...
- Codeforces Round #290 (Div. 2) D. Fox And Jumping dp
D. Fox And Jumping 题目连接: http://codeforces.com/contest/510/problem/D Description Fox Ciel is playing ...
- Codeforces Round #290 (Div. 2) C. Fox And Names dfs
C. Fox And Names 题目连接: http://codeforces.com/contest/510/problem/C Description Fox Ciel is going to ...
- Codeforces Round #290 (Div. 2) A. Fox And Snake 水题
A. Fox And Snake 题目连接: http://codeforces.com/contest/510/problem/A Description Fox Ciel starts to le ...
- Codeforces Round #290 (Div. 2) E. Fox And Dinner 网络流建模
E. Fox And Dinner time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- 找规律 Codeforces Round #290 (Div. 2) A. Fox And Snake
题目传送门 /* 水题 找规律输出 */ #include <cstdio> #include <iostream> #include <cstring> #inc ...
- 拓扑排序 Codeforces Round #290 (Div. 2) C. Fox And Names
题目传送门 /* 给出n个字符串,求是否有一个“字典序”使得n个字符串是从小到大排序 拓扑排序 详细解释:http://www.2cto.com/kf/201502/374966.html */ #i ...
- CodeForces Round #290 Div.2
A. Fox And Snake 代码可能有点挫,但能够快速A掉就够了. #include <cstdio> int main() { //freopen("in.txt&quo ...
随机推荐
- OE7设置菜单为什么这么少?
默认安装的OE7设置菜单只有很少的功能: 如果需要更多的OE定制,必须开启“技术特性”选项:
- 关于Cygwin的x-Server的自动运行以及相关脚本修改
常常需要用到远端服务器的图形工具,如果在windows端没用xserver的话,很多程序无法运行.一个特殊的例子,emacs在没用xserver的时候,是直接在终端中打开的,如果不修改cygwin.b ...
- 事务处理: databse jdbc mybatis spring
事务的认识需要一个相当漫长的流程,慢慢在实践中理解,然后在强化相关理论基础. 数据库中的事务: 传统的本地事务处理都是依靠数据库自身事务处理能力,而事务本身是传统关系型数据库的基石.简单来说事务就是一 ...
- 写在阿里去IOE一周年
[文/ 任英杰] 去年5月17日,阿里巴巴支付宝最后一台IBM小型机在下线,标志着阿里完成去IOE.随后一场去IOE运动不断发酵,甚至传闻IBM中国去年损失了20%的合同额. 去了IOE,奔向何方?阿 ...
- 闲置小U盘变身最强大路由器
小容量 U 盘,用起来嫌容量太少,丢了好像又觉得太可惜.不过现在将它进行一番小改造后,配合我们的电脑 ,就能得到一台强大的路由器,不仅省了买路由的钱,而且这台路由器在市面上基本买不到 ! DD ...
- Weibo Crawler in Action
1.要写一个微博爬虫,得分开几个模块来做: (1)模拟登录 (2)模拟浏览 (3)针对短时间内大量访问而引起怀疑的禁止登陆解决方案 (4)其他 (1)模拟登陆模块 前提:要模拟登录,得首先知道在登录微 ...
- iOS OpenCV 缺少64位解决方法
- delphi请求idhttp数据
idhttp ss : TStringStream; begin ss := TStringStream.)); { 指定gb2312的中文代码页,或者54936(gb18030)更好些 utf8 对 ...
- java.lang.IllegalStateException: Cannot call sendError() after the response has been committed
http://blog.csdn.net/chenghui0317/article/details/9531171 —————————————————————————————————————————— ...
- 1.VS2010C++环境设置
一.需要下载的软件 1.visual studio 2010\\xxzx\tools\编程工具\MICROSOFT\VISUAL.STUDIO\VISUAL.STUDIO.201032位cn_visu ...