C. White Sheet

There is a white sheet of paper lying on a rectangle table. The sheet is a rectangle with its sides parallel to the sides of the table. If you will take a look from above and assume that the bottom left corner of the table has coordinates (0,0), and coordinate axes are left and bottom sides of the table, then the bottom left corner of the white sheet has coordinates (x1,y1), and the top right — (x2,y2).

After that two black sheets of paper are placed on the table. Sides of both black sheets are also parallel to the sides of the table. Coordinates of the bottom left corner of the first black sheet are (x3,y3), and the top right — (x4,y4). Coordinates of the bottom left corner of the second black sheet are (x5,y5), and the top right — (x6,y6).

Example of three rectangles.
Determine if some part of the white sheet can be seen from the above after the two black sheets are placed. The part of the white sheet can be seen if there is at least one point lying not strictly inside the white sheet and strictly outside of both black sheets.

Input
The first line of the input contains four integers x1,y1,x2,y2 (0≤x1<x2≤106,0≤y1<y2≤106) — coordinates of the bottom left and the top right corners of the white sheet.

The second line of the input contains four integers x3,y3,x4,y4 (0≤x3<x4≤106,0≤y3<y4≤106) — coordinates of the bottom left and the top right corners of the first black sheet.

The third line of the input contains four integers x5,y5,x6,y6 (0≤x5<x6≤106,0≤y5<y6≤106) — coordinates of the bottom left and the top right corners of the second black sheet.

The sides of each sheet of paper are parallel (perpendicular) to the coordinate axes.

Output
If some part of the white sheet can be seen from the above after the two black sheets are placed, print "YES" (without quotes). Otherwise print "NO".

Input


Output

NO

Input


Output

YES

思路:判断白色矩阵是否四个顶点都在黑色矩阵内,如果不是,直接YES;否则,判断是否白色矩阵被某一个黑色矩阵包含,是就NO,不是的话判断两个黑色矩阵是否相离,是就NO,不是就YES;【借鉴博客的.】

AC代码:

 #include<bits/stdc++.h>

 using namespace std;

 int x1,x2,x3,x4,x5,x6;
int y1,y2,y3,y4,y5,y6;
int check(int x,int y){
int flag=;
if(x>=x3&&x<=x4&&x>=y3&&x<=y4)
flag=;
if(flag){
return flag;
}
if(x>=x5&&x<=x6&&x>=y5&&x<=y6)
flag=;
return flag;
}
int check1(){
int flag=;
if(x1>=x3&&y1>=y3&&x2<=x4&&y2<=y4){
flag=;
}
return flag;
}
int check2(){
int flag=;
if(x1>=x5&&y1>=y5&&x2<=x6&&y2<=y6){
flag=;
}
return flag;
}
int main(){
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
scanf("%d%d%d%d",&x3,&y3,&x4,&y4);
scanf("%d%d%d%d",&x5,&y5,&x6,&y6);
int a=check(x1,y1);
int b=check(x1,y2);
int c=check(x2,y2);
int d=check(x2,y1);
if(!a||!b||!c||!d){ // 判断是否四边形是否有点没有被覆盖
printf("YES");
return ;
}
if(check1()||check2()){ // 判断是否有矩形可将其完全覆盖的
printf("NO");
return ;
}
if(max(x3,x5)>min(x4,x6)||max(y3,y5)>min(y4,y6)){ // 判断是否矩形分离
printf("YES");
}else{
printf("NO");
}
return ;
}

Codeforces Round #587 (Div. 3) C题 【判断两个矩形是否完全覆盖一个矩形问题】 {补题 [差点上分系列]}的更多相关文章

  1. Codeforces Round #587 (Div. 3) C. White Sheet

    链接: https://codeforces.com/contest/1216/problem/C 题意: There is a white sheet of paper lying on a rec ...

  2. Codeforces Round #587 (Div. 3)

    https://codeforces.com/contest/1216/problem/A A. Prefixes 题意大概就是每个偶数位置前面的ab数目要相等,很水,被自己坑了 1是没看见要输出修改 ...

  3. Codeforces Round #324 (Div. 2) (快速判断素数模板)

    蛋疼的比赛,当天忘了做了,做的模拟,太久没怎么做题了,然后C题这么简单的思路却一直卡到死,期间看了下D然后随便猜了下,暴力了下就过了. A.找一个能被t整除的n位数,那么除了<=10以外,其他都 ...

  4. Codeforces Round #587 (Div. 3) D. Swords

    链接: https://codeforces.com/contest/1216/problem/D 题意: There were n types of swords in the theater ba ...

  5. Codeforces Round #587 (Div. 3) B. Shooting(贪心)

    链接: https://codeforces.com/contest/1216/problem/B 题意: Recently Vasya decided to improve his pistol s ...

  6. Codeforces Round #587 (Div. 3) A. Prefixes

    链接: https://codeforces.com/contest/1216/problem/A 题意: Nikolay got a string s of even length n, which ...

  7. Codeforces Round #587 (Div. 3) F. Wi-Fi(单调队列优化DP)

    题目:https://codeforces.com/contest/1216/problem/F 题意:一排有n个位置,我要让所有点都能联网,我有两种方式联网,第一种,我直接让当前点联网,花费为i,第 ...

  8. Codeforces Round #452 (Div. 2)-899A.Splitting in Teams 899B.Months and Years 899C.Dividing the numbers(规律题)

    A. Splitting in Teams time limit per test 1 second memory limit per test 256 megabytes input standar ...

  9. Codeforces Round #587 (Div. 3) F Wi-Fi(线段树+dp)

    题意:给定一个字符串s 现在让你用最小的花费 覆盖所有区间 思路:dp[i]表示前i个全覆盖以后的花费 如果是0 我们只能直接加上当前位置的权值 否则 我们可以区间询问一下最小值 然后更新 #incl ...

随机推荐

  1. jenkins sonarqube 代码检测部署

    install pgsql and sonarqube docker run --name postgresqldb -e POSTGRES_USER=sonar -e POSTGRES_PASSWO ...

  2. 切片(Slice)

    Python提供了切片(Slice)操作符:可以一次取出多个列表元素 L[0:3]表示,从索引0开始取,直到索引3为止,但不包括索引3.0可以省略:L[:3] L[:]:就是整个列表   补充: 前1 ...

  3. 防止vi粘贴时自动添加缩进的方法

    使用Xshell连接Linux服务器,使用vi打开文件进行粘贴时,会自动在行首添加很多空格,导致格式错乱.可以用如下方法剞劂 在拷贝前输入:set paste (这样的话,vim就不会启动自动缩进,而 ...

  4. Manacher算法+注释

    Manacher算法是用来求一个字符串中最长回文串的算法. 考虑暴力求最长回文串的做法: 暴力枚举字符串中的所有字串判断是否回文,然后求最大值. 时间复杂度O(n^3),考虑优化. 我们从枚举所有字串 ...

  5. mysql 树结构递归处理

    日常开发中我们经常会遇到树形结构数据处理,一般表结构通常会常用id,pid这种设计方案. 之前用oracle.sqlServer数据库,用相应的语法即可获取树形结构数据(oracel:connect ...

  6. python 切换虚拟环境

    每次电脑重启后,都要切入虚拟环境,命令总是忘记.如果使用IDE,可以指定interpeter方便的切换. 首先conda info --env  查看当前有几个环境 激活/禁用环境 source ac ...

  7. ubuntu14.04 caffe

    1.显卡驱动 ubuntu nvidia 940m 使用sudo ubuntu-drivers devices 查看推荐的驱动版本 //sudo add-apt-repository ppa:mama ...

  8. ThinkPad T420i 上 Ubuntu 12.04 实现指纹识别登录

    ThinkPad T420i 上 Ubuntu 12.04 实现指纹识别登录 # add ppa add-apt-repository ppa:fingerprint/fprint # update ...

  9. 1 使用webpack搭建vue开发环境

    1 先去node.js官网下载nodejs并且安装 安装成功之后在命令行输入node -v 回车,npm -v回车如果显示对应的版本号,说明node安装成功,自带的npm也安装成功 2 在d盘下创建一 ...

  10. C++ 语句函数再探

    1. 表达式只计算,抛弃计算结果: 2. 空语句什么也不做: 3.switch case语句漏写break,将会从匹配到的情况开始执行,直到语句结束 int main() { ; i + ; //表达 ...