Vitaly works at the warehouse. The warehouse can be represented as a grid of n × m cells, each of which either is free or is occupied by a container. From every free cell it's possible to reach every other free cell by moving only through the cells sharing a side. Besides that, there are two robots in the warehouse. The robots are located in different free cells.

Vitaly wants to swap the robots. Robots can move only through free cells sharing a side, moreover, they can't be in the same cell at the same time or move through each other. Find out if the swap can be done.

Input

The first line contains two positive integers n and m (2 ≤ n·m ≤ 200000) — the sizes of the warehouse.

Each of the next n lines contains m characters. The j-th character of the i-th line is «.» if the corresponding cell is free, «#» if there is a container on it, «1» if it's occupied by the first robot, and «2» if it's occupied by the second robot. The characters «1» and «2» appear exactly once in these lines.

Output

Output «YES» (without quotes) if the robots can be swapped, and «NO» (without quotes) if that can't be done.

Examples
Input
5 3

###

#1#

#.#

#2#

###
Output
NO
Input
3 5

#...#

#1.2#

#####
Output
YES
bfs标记数组的应用
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#pragma GCC diagnostic error "-std=c++11"
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int dir[][]={{,},{,-},{-,},{,}};
int main()
{
int n,m,pos_one_x,pos_one_y,pos_two_x,pos_two_y;
scanf("%d%d",&n,&m);
char ch[n+][m+];
int vis[n+][m+],ans[n+][m+],tot=;
memset(ans,,sizeof(ans));
for(int i=;i<=n;i++)
{
scanf("%s",ch[i]+);
for(int j=;j<=m;j++)
{
if(ch[i][j]=='') pos_one_x=i,pos_one_y=j;
if(ch[i][j]=='') pos_two_x=i,pos_two_y=j;
}
}
memset(vis,,sizeof(vis));
queue<pair<int,int> >q;
vis[pos_one_x][pos_one_y]=;
q.push({pos_one_x,pos_one_y});
while(!q.empty())
{
pair<int,int> p=q.front();
q.pop();
for(int i=;i<;i++)
{
int xx=p.first+dir[i][];
int yy=p.second+dir[i][];
if(xx> && xx<=n && yy> && yy<=m && !vis[xx][yy] && ch[xx][yy]!='#')
{
vis[xx][yy]=;
q.push({xx,yy});
}
}
}
if(!vis[pos_two_x][pos_two_y]) return *printf("NO\n");
bool flag=false;
for(int i=;i<=n;i++)
for(int j=;j<=m;j++)
{
if(vis[i][j])
{
if(vis[i-][j]) ans[i][j]++;
if(vis[i+][j]) ans[i][j]++;
if(vis[i][j-]) ans[i][j]++;
if(vis[i][j+]) ans[i][j]++;
if(ans[i][j]>=) flag=true;
if(ans[i][j]==) tot++;
}
}
if(tot!=) flag=true;
puts(flag?"YES":"NO");
return ;
}

gym 100971 J Robots at Warehouse的更多相关文章

  1. 【Gym 100971J】Robots at Warehouse

    题意 链接给你一个n*m的地图,'#'代表墙,‘.’代表可走的,1代表1号机器人,2代表2号机器人,机器人可以上下左右移动到非墙的位置,但不能走到另一个机器人身上.问能否交换1和2的位置. 分析 如果 ...

  2. codeforces gym 100971 K Palindromization 思路

    题目链接:http://codeforces.com/gym/100971/problem/K K. Palindromization time limit per test 2.0 s memory ...

  3. Codeforces gym 101343 J.Husam and the Broken Present 2【状压dp】

     2017 JUST Programming Contest 2.0 题目链接:Codeforces gym 101343 J.Husam and the Broken Present 2 J. Hu ...

  4. 【codeforces.com/gym/100240 J】

    http://codeforces.com/gym/100240 J [分析] 这题我搞了好久才搞出样例的11.76....[期望没学好 然后好不容易弄成分数形式.然后我‘+’没打..[于是爆0... ...

  5. Codeforces GYM 100876 J - Buying roads 题解

    Codeforces GYM 100876 J - Buying roads 题解 才不是因为有了图床来测试一下呢,哼( 题意 给你\(N\)个点,\(M\)条带权边的无向图,选出\(K\)条边,使得 ...

  6. codeforces Gym 100187J J. Deck Shuffling dfs

    J. Deck Shuffling Time Limit: 2   Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/pro ...

  7. codeforces GYM 100114 J. Computer Network 无相图缩点+树的直径

    题目链接: http://codeforces.com/gym/100114 Description The computer network of “Plunder & Flee Inc.” ...

  8. codeforces Gym 100500 J. Bye Bye Russia

    Problem J. Bye Bye RussiaTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/1005 ...

  9. codeforces GYM 100114 J. Computer Network tarjan 树的直径 缩点

    J. Computer Network Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Des ...

随机推荐

  1. poj1363——Rails

    Description There is a famous railway station in PopPush City. Country there is incredibly hilly. Th ...

  2. android sudio 执行的中文是乱码解决方式

    1.File-->Setings-->查找file encodings 例如以下图 2.将 IDE Encoding .Project Encoding.Default encoding ...

  3. Windows环境下通过Git来管理自己的Android代码

    前面已经介绍了在Windows下使用git工具来下载Android的源代码,Windows环境下通过Git得到Android源代码,这里记录我使用git工具来管理我自己的代码,git是一种分布式的项目 ...

  4. 火狐访问IIS出现404,而Chrome可以正常访问

    需要在web.config中的handlers中添加如下节点,保存之后,需要重启电脑. <remove name="ExtensionlessUrlHandler-Integrated ...

  5. JAVA设计模式之【适配器模式】

    适配器模式 当不需要实现一个接口所提供的所有方法时,可先设计一个抽象类该接口,并为接口每个方法提供一个默认实现 该抽象类的子类可以选择性地覆盖父类的某些方法来实现需求 角色 适配者接口 通常在接口中声 ...

  6. nyoj--55--懒省事的小明(STL优先队列)

    懒省事的小明 时间限制:3000 ms  |  内存限制:65535 KB 难度:3 描述       小明很想吃果子,正好果园果子熟了.在果园里,小明已经将所有的果子打了下来,而且按果子的不同种类分 ...

  7. requireJS实现原理分析

    下面requireJS实现的基本思路  项目地址https://github.com/WangMaoling/require var require = (function(){ //框架版本基本信息 ...

  8. ibatis annotations 注解方式返回刚插入的自增长主键ID的值--转

    原文地址:http://www.blogs8.cn/posts/WWpt35l mybatis提供了注解方式编写sql,省去了配置并编写xml mapper文件的麻烦,今天遇到了获取自增长主键返回值的 ...

  9. 机器学习(十一) 支持向量机 SVM(上)

    一.什么是支撑向量机SVM (Support Vector Machine) SVM(Support Vector Machine)指的是支持向量机,是常见的一种判别方法.在机器学习领域,是一个有监督 ...

  10. 转js resplace方法使用

    作者: hezhiwu5#163.com    时间:2007-3-22 大家好!!今晚在华软G43*宿舍没什么事做,把javascript中replace方法讲解一下,如果讲得不对或不合理是情理之中 ...