Codeforces812A Sagheer and Crossroads 2017-06-02 20:41 139人阅读 评论(0) 收藏
1 second
256 megabytes
standard input
standard output
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes
getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts
in total. Each part has 4 lights, one for each lane getting into the intersection (l —
left, s — straight, r —
right) and a light p for a pedestrian crossing.
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p —
for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for
green light.
On a single line, print "YES" if an accident is possible, and "NO"
otherwise.
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
YES
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
NO
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
NO
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4.
Also, cars of parts 2 and 3can
hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath> using namespace std; #define LL long long
const int inf=0x7fffffff; int main()
{
int a[10][10];
for(int i=1; i<=4; i++)
for(int j=1; j<=4; j++)
scanf("%d",&a[i][j]);
int flag=0;
if(a[1][4]==1&&(a[1][1]==1||a[1][2]==1||a[1][3]==1||a[2][1]==1||a[3][2]==1||a[4][3]==1))
printf("YES\n"),flag=1;
else if(a[2][4]==1&&(a[2][1]==1||a[2][2]==1||a[2][3]==1||a[3][1]==1||a[4][2]==1||a[1][3]==1))
printf("YES\n"),flag=1;
else if(a[3][4]==1&&(a[3][1]==1||a[3][2]==1||a[3][3]==1||a[4][1]==1||a[1][2]==1||a[2][3]==1))
printf("YES\n"),flag=1;
else if(a[4][4]==1&&(a[4][1]==1||a[4][2]==1||a[4][3]==1||a[1][1]==1||a[2][2]==1||a[3][3]==1))
printf("YES\n"),flag=1;
if(flag==0)
printf("NO\n"); return 0;
}
Codeforces812A Sagheer and Crossroads 2017-06-02 20:41 139人阅读 评论(0) 收藏的更多相关文章
- python如何使用 os.path.exists()--Learning from stackoverflow 分类: python 2015-04-23 20:48 139人阅读 评论(0) 收藏
Q&A参考连接 Problem:IOError: [Errno 2] No such file or directory. os.path.exists() 如果目录不存在,会返回一个0值. ...
- Codeforces812B Sagheer, the Hausmeister 2017-06-02 20:47 85人阅读 评论(0) 收藏
B. Sagheer, the Hausmeister time limit per test 1 second memory limit per test 256 megabytes input s ...
- Codeforces812C Sagheer and Nubian Market 2017-06-02 20:39 153人阅读 评论(0) 收藏
C. Sagheer and Nubian Market time limit per test 2 seconds memory limit per test 256 megabytes input ...
- hadoop调优之一:概述 分类: A1_HADOOP B3_LINUX 2015-03-13 20:51 395人阅读 评论(0) 收藏
hadoop集群性能低下的常见原因 (一)硬件环境 1.CPU/内存不足,或未充分利用 2.网络原因 3.磁盘原因 (二)map任务原因 1.输入文件中小文件过多,导致多次启动和停止JVM进程.可以设 ...
- Self Numbers 分类: POJ 2015-06-12 20:07 14人阅读 评论(0) 收藏
Self Numbers Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 22101 Accepted: 12429 De ...
- Debian自启动知识 2015-03-31 20:23 79人阅读 评论(0) 收藏
Debian6添加了insserv用来代替update-rc.d.update-rc.d 就不多做介绍. Debian6里边要添加一个自动启动的服务需要先将启动脚本放在/etc/init.d,然后使用 ...
- UI基础:UIView(window,frame,UIColor,CGPoint,alpha,CGRect等) 分类: iOS学习-UI 2015-06-30 20:01 119人阅读 评论(0) 收藏
UIView 视图类,视图都是UIView或者UIView子类 UIWindow 窗口类,用于展示视图,视图一定要添加window才能显示 注意:一般来说,一个应用只有一个window 创建一个UIW ...
- OC基础:OC 基本数据类型与对象之间的转换方法 分类: ios学习 OC 2015-06-18 20:01 11人阅读 评论(0) 收藏
1.Foundation框架中提供了很多的集合类如:NSArray,NSMutableArray,NSSet,NSMutableSet,NSDictionary,NSMutableDictionary ...
- ZOJ2748 Free Kick 2017-04-18 20:40 40人阅读 评论(0) 收藏
Free Kick Time Limit: 2 Seconds Memory Limit: 65536 KB In a soccer game, a direct free kick is ...
随机推荐
- vs 调试 IE显示“无法显示该网页
在用VS2010调试网站的时候,突然页面不能正常显示了,IE显示“无法显示该网页”.症状一: IE地址栏里面显示的端口号和桌面任务栏右下角“ASP.NET Development Server”的端口 ...
- java有车有房有能力最基本运用
public class yunsuan { public static void main(String[] args) { // 1是有,0是没有 int i = 1, l = 0;// 有房 i ...
- 把一行数字(readline)读进List并以科学计数法输出(write)到文件
主要过程是读取的时候是一行字符串,需要Strip去除空格等,然后split变成一个List. 注意这时候数据结构是List但是每一个元素是Str性质的. 所以需要map(float,List) 把这 ...
- c中extern的作用
参考资料: http://www.cnblogs.com/yc_sunniwell/archive/2010/07/14/1777431.html
- .NET获取城市信息(将三字代码转换成城市名)
整理代码,发现有一个从两张表里读取城市列表,然后linq和lambda表达式来获取城市名的函数,代码如下: public static string GetCityHotelText(string c ...
- 利用PHP脚本辅助MySQL数据库管理1-表结构
<?php $dbi = new DbMysql; $dbi->dbh = 'mysql://root:mysql@127.0.0.1/coffeetest'; $map = array( ...
- 买铅笔(NOIP2016)
先给题目链接:买铅笔 这题非常水,没啥可分析的,先给代码: #include<bits/stdc++.h> //1 using namespace std; int main(){ int ...
- ServiceDesk Plus 服务管理自动指派工单功能
- Hibernate 的缓存
Hibernate的一级缓存 什么是缓存?? 1 数据存到数据库里面,数据库本身是文件系统,使用流方式操作文件效率不是很高. 1.1 把数据存到内存里面,不需要使用流方式,可以直接读取内存中数据 ...
- 2019.01.14 codeforces685B. Kay and Snowflake(树形dp)
传送门 题意简述:给出一棵树,求每个子树的重心. 首先通过画图可以观察出一个性质,我们从叶子结点向根节点递推重心的话重心的位置是不会下降的. 然后由于一个点的重心要么是自己,要么在重儿子子树内,因此如 ...