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
- mysql mysqld.sock文件丢失问题
修改mysql 编码为utf8时 在/etc/mysql/目录下 在 [client] 添加 default-character-set=utf8 [mysqld]添加 default-charact ...
- 案例情景--在一次Oracle 数据库导出时 EXP-00008;ORA-00904:EXP-00000: oracle不同版本导入导出规则
案例情景--在一次Oracle 数据库导出时: C:\Documents and Settings\Administrator>exp lsxy/lsxy@lsxy_db file=E:\lsx ...
- POJ3662或洛谷1948 Telephone Lines
二分答案+单源最短路 POJ原题链接 洛谷原题链接 显然可以二分答案,检验\(mid\)可以使用最短路来解决. 将大于\(mid\)的边看成长度为\(1\)的边,说明要使用免费升级服务,否则长度为\( ...
- ZBlog你选择PHP还是ASP?
最近趁着空闲玩了一下zblog,对于很多第一次接触zblog的博主大多都会问zblog是PHP好还是ASP好?我们应该如何选择?其实,对于这个问题我也不是很懂,我个人比较倾向于PHP.今天我就整理一下 ...
- @1-2初识Python爬虫
初识Python爬虫 Python爬虫(入门+进阶) DC学院 环境搭建: Python2与Python3的差异:python2与python3整体差异不大,大多是一些语法上的区别,考虑到py ...
- 浅谈Spring中的Quartz配置
浅谈Spring中的Quartz配置 2009-06-26 14:04 樊凯 博客园 字号:T | T Quartz是一个强大的企业级任务调度框架,Spring中继承并简化了Quartz,下面就看看在 ...
- 超高清视频会议所需带宽分析---1M带宽应用720P是否可能?
超高清视频会议所需带宽分析 ---1M带宽应用720P是否可能? 1.首先计算720P(1280×720)单幅图像照片的数据量 每像素用24比特表示,则:720P图像照片的原始数据量= 1280×72 ...
- PHP字符串转实体函数
与HTML实体相关的函数 htmlspecialchars函数 描述:预定义的字符转换为HTML实体 语法:string htmlspecialchars(string $string [,int $ ...
- python学习 day15 (3月20日)----time
# '2019-03-20 10:40:00'#这个时间向后推一个月 f1 = time.strptime('2019-03-20 10:40','%Y-%m-%d %H:%M') # 把字符串时间转 ...
- 693. Binary Number with Alternating Bits
static int wing=[]() { std::ios::sync_with_stdio(false); cin.tie(NULL); ; }(); class Solution { publ ...