(中等) CF 585B Phillip and Trains,BFS。
The mobile application store has a new game called "Subway Roller".
The protagonist of the game Philip is located in one end of the tunnel and wants to get out of the other one. The tunnel is a rectangular field consisting of three rows and n columns. At the beginning of the game the hero is in some cell of the leftmost column. Some number of trains rides towards the hero. Each train consists of two or more neighbouring cells in some row of the field.
All trains are moving from right to left at a speed of two cells per second, and the hero runs from left to right at the speed of one cell per second. For simplicity, the game is implemented so that the hero and the trains move in turns. First, the hero moves one cell to the right, then one square up or down, or stays idle. Then all the trains move twice simultaneously one cell to the left. Thus, in one move, Philip definitely makes a move to the right and can move up or down. If at any point, Philip is in the same cell with a train, he loses. If the train reaches the left column, it continues to move as before, leaving the tunnel.
Your task is to answer the question whether there is a sequence of movements of Philip, such that he would be able to get to the rightmost column.
题意就是问能不能从左到右,假设车不动,然后人向右走,一次一步,分析可知只有当人在的行数是mod 3=2时才能向上下,否则只能向右。
所以就可以直接BFS。
但是还要注意一个问题就是如果上一步是上下那么这一步就不能是,因为这个被hack了,好不爽。。。
代码如下:
// ━━━━━━神兽出没━━━━━━
// ┏┓ ┏┓
// ┏┛┻━━━━━━━┛┻┓
// ┃ ┃
// ┃ ━ ┃
// ████━████ ┃
// ┃ ┃
// ┃ ┻ ┃
// ┃ ┃
// ┗━┓ ┏━┛
// ┃ ┃
// ┃ ┃
// ┃ ┗━━━┓
// ┃ ┣┓
// ┃ ┏┛
// ┗┓┓┏━━━━━┳┓┏┛
// ┃┫┫ ┃┫┫
// ┗┻┛ ┗┻┛
//
// ━━━━━━感觉萌萌哒━━━━━━ // Author : WhyWhy
// Created Time : 2015年10月12日 星期一 18时20分55秒
// File Name : D.cpp #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h> using namespace std; const int MaxN=; int N,K; char map1[][MaxN];
int vis[][MaxN]; int que[];
int first,last; void bfs()
{
first=last=;
int p;
int x,y;
int t;
int u;
for(int i=;i<=;++i) if(map1[i][]=='s') p=i; que[last++]=p*+;
vis[p][]=; while(last-first)
{
u=que[first++];
x=u/;
y=u%;
t=vis[x][y]; if(y<N && vis[x][y+]!= && map1[x][y+]=='.')
{
vis[x][y+]=;
que[last++]=x*+y+;
} if(y%== && x> && t== && vis[x-][y]== && map1[x-][y]=='.')
{
vis[x-][y]=;
que[last++]=(x-)*+y;
} if(y%== && x< && t== && vis[x+][y]== && map1[x+][y]=='.')
{
vis[x+][y]=;
que[last++]=(x+)*+y;
}
}
} int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout); int T;
scanf("%d",&T); while(T--)
{
scanf("%d %d",&N,&K);
for(int i=;i<=;++i)
scanf("%s",map1[i]+);
memset(vis,,sizeof(vis)); bfs(); if(vis[][N] || vis[][N] || vis[][N])
puts("YES");
else puts("NO");
} return ;
}
(中等) CF 585B Phillip and Trains,BFS。的更多相关文章
- Codeforces Round #325 (Div. 2) D. Phillip and Trains BFS
D. Phillip and Trains Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/586/ ...
- Codeforces 586D. Phillip and Trains 搜索
D. Phillip and Trains time limit per test: 1 second memory limit per test :256 megabytes input: stan ...
- Codeforces 586D Phillip and Trains(DP)
题目链接 Phillip and Trains 考虑相对位移. 每一轮人向右移动一格,再在竖直方向上移动0~1格,列车再向左移动两格. 这个过程相当于每一轮人向右移动一格,再在竖直方向上移动0~1格, ...
- CF586D. Phillip and Trains
/* CF586D. Phillip and Trains http://codeforces.com/problemset/problem/586/D 搜索 */ #include<cstdi ...
- CF #375 (Div. 2) D. bfs
1.CF #375 (Div. 2) D. Lakes in Berland 2.总结:麻烦的bfs,但其实很水.. 3.题意:n*m的陆地与水泽,水泽在边界表示连通海洋.最后要剩k个湖,总要填掉多 ...
- CF 520 B. Two Buttons(bfs)
/*题意:一个数,就是输入的第一个数,让它变成第二个数最少用几步.可以点红色按钮,蓝色按钮来改变数字,红色:*2,蓝色:-1,如果变成负数,就变成原来的数.CF 520 B. Two Buttons思 ...
- 【33.33%】【codeforces 586D】Phillip and Trains
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- cf.295.B Two Buttons (bfs)
Two Buttons time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...
- (中等) CF 555E Case of Computer Network,双连通+树。
Andrewid the Android is a galaxy-known detective. Now he is preparing a defense against a possible a ...
随机推荐
- 学习最短路建图 HUD 5521
http://acm.hdu.edu.cn/showproblem.php?pid=5521 题目大意:有n个点,m个集合,每个集合里面的点都两两可达且每条边权值都是val,有两个人A, B,A在po ...
- mongodb分片
在系统早期,数据量还小的时候不会引起太大的问题,但是随着数据量持续增多,后续迟早会出现一台机器硬件瓶颈问题的.而mongodb主打的就是海量数据架构,他不能解决海量数据怎么行!不行!“分片”就用这个来 ...
- androidstudio 问题
Error:(1, 1) A problem occurred evaluating project ':app'. > Failed to apply plugin [id 'com.andr ...
- sudo su– user
[root@localhost ~] # visudo –f /etc/sudoers 在文件中的root账户下添加需要切换root账户的账户 root ALL=(ALL) ALL user ALL= ...
- List<T>转换为ObservableCollection<T>
ObservableCollection能通知他变化了也正是因为它实现了INotifyPropertyChanged接口, 在wpf项目中经常会遇到把List<T>转换为Observabl ...
- linux cat more less head tail
cat 命令: cat filename 查看一个文件的内容cat[选项][文件]... -b 对非空白行进行编号,行号从1开始-n 和nl命令差不多,对所有行(包括空白行)进行编号输出显示-E ...
- 软件开发常用的linux命令心得(ubuntu为例)
软件开发过程中难免要经常对主机进行配置或者部署等操作,想到一些就写一些了,以后再更新 解压命令: a.如果是tar文件,则直接用 “tar zxvf 文件名”: b.如果是zip文件,用 “unzip ...
- ACM课程学习总结
ACM课程学习总结报告 通过一个学期的ACM课程的学习,我学习了到了许多算法方面的知识,感受到了算法知识的精彩与博大,以及算法在解决问题时的巨大作用.此篇ACM课程学习总结报告将从以下方面展开: 学习 ...
- DevExpress.LookUpEdit控件实现自动搜索定位功能 兼使用方法(looUpEdit可编辑)
DevExpress.LookUpEdit 使用方法 设置可手动输入 this.LookUpEdit1.Properties.TextEditStyle = DevExpress.XtraEditor ...
- hadoop的安全模式
在安全模式下:不能增.删.改操作:但可以查看. 查看hadoop是否i处于安全模式下: 执行命令:hadoop dfsadmin -safemode get 进入hadoop的安全模式下: 执行命令: ...