gym 100971 J Robots at Warehouse
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.
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 «YES» (without quotes) if the robots can be swapped, and «NO» (without quotes) if that can't be done.
5 3 ### #1# #.# #2# ###
NO
3 5 #...# #1.2# #####
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
- 【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 ...
- 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 ...
- 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 ...
- 【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... ...
- 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 ...
- 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 ...
- 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.” ...
- 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 ...
- 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
- [poj3974] Palindrome 解题报告 (hash\manacher)
题目链接:http://poj.org/problem?id=3974 题目: 多组询问,每组给出一个字符串,求该字符串最长回文串的长度 数据范围支持$O(nlog n)$ 解法一: 二分+hash ...
- WPF学习(三) - 依赖属性
学习WPF时,我在看一本叫做“深入浅出WPF”的书.整整20页都在讲依赖性性和附加属性,反复看了几遍居然还是不懂,真是郁闷. 上一篇中WPF绑定的例子,其实已经用到了依赖属性. // 作为被绑定的目标 ...
- 你不知道的JavaScript(十一)函数参数
JavaScript函数的定义和使用非常简单,我们看一下下面的代码: <script type="text/javascript"> var sum = functio ...
- 新疆大学(新大)OJ xju 1006: 比赛排名 第二类斯特林数+阶乘
题目链接:http://acm.xju.edu.cn/JudgeOnline/problem.php?id=1006 第二类斯特林数: 第二类Stirling数实际上是集合的一个拆分,表示将n个不同的 ...
- 51nod 1402 最大值 3级算法题 排序后修改限制点 时间复杂度O(m^2)
代码: 题意,第一个数为0,相邻的数相差0或者1,有一些点有限制,不大于给定值,求这组数中可能的最大的那个数. 这题我们看一个例子:第5个数的限制为2 1 2 3 4 5 6 7 8 9 0 1 2 ...
- Servlet学习(四)——response
1.概述 在创建Servlet时会覆盖service()方法,或doGet()或doPost(),这些方法都有两个参数,一个是代表请求的request和代表响应response. service方法中 ...
- CorelDRAW X6低价再次冲破底线
平时我们看到的标志设计.杂志排版.产品商标.插图描画......这些都是设计师们使用CorelDRAW设计而来.如今CorelDRAW已经成为每个设计师必装的软件,从12年发布CorelDRAW X6 ...
- Set集合[HashSet,TreeSet,LinkedHashSet],Map集合[HashMap,HashTable,TreeMap]
------------ Set ------------------- 有序: 根据添加元素顺序判定, 如果输出的结果和添加元素顺序是一样 无序: 根据添加元素顺序判定,如果输出的结果和添加元素的顺 ...
- eclipse的maven工程视图切换
上面图切换成下面图: 点击eclipse右上角,如下图红圈,然后在选择javaEE这样就切换成javaEE视图了
- [ZJOI2006]物流运输 最短路 动态规划
Code: 定义状态 $dp[i]$ 为前 $i$ 天的最小代价. 状态转移为:$dp[i]=min(dp[i],dp[j]+spfa(j+1,i)$ 这里 $spfa(i,j)$ 是指 $(i,j) ...