Calabash and Landlord

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3228    Accepted Submission(s): 613

Problem Description
Calabash is the servant of a landlord. The landlord owns a piece of land, which can be regarded as an infinite 2D plane.

One
day the landlord set up two orthogonal rectangular-shaped fences on his
land. He asked Calabash a simple problem: how many nonempty connected
components is my land divided into by these two fences, both finite and
infinite? Calabash couldn't answer this simple question. Please help
him!

Recall that a connected component is a maximal set of
points not occupied by the fences, and every two points in the set are
reachable without crossing the fence.

 
Input
The first line of input consists of a single integer T (1≤T≤10000), the number of test cases.

Each test case contains two lines, specifying the two rectangles. Each line contains four integers x1,y1,x2,y2 (0≤x1,y1,x2,y2≤109,x1<x2,y1<y2), where (x1,y1),(x2,y2)
are the Cartesian coordinates of two opposite vertices of the
rectangular fence. The edges of the rectangles are parallel to the
coordinate axes. The edges of the two rectangles may intersect, overlap,
or even coincide.

 
Output
For each test case, print the answer as an integer in one line.
 
Sample Input
3
0 0 1 1
2 2 3 4
1 0 3 2
0 1 2 3
0 0 1 1
0 0 1 1
 
Sample Output
3
4
2
 

题意:给你2个栅栏的坐标,栅栏会分割平面,问栅栏分割平面后有多少个连通块

思路:枚举所有情况,时间复杂度O(1),代码复杂度O(∞),一开始WA到自闭,就开始暴力模式,写的时候懵了,出现了一些难以发现的小错误,差点翻车,最后15分钟才改出来,

所有情况都在代码里了,可以看一下代码,这里就不细讲了

 #include<bits/stdc++.h>
using namespace std;
int x_1[],y_1[],x_2[],y_2[];
int main(){
int t,a,b;
ios::sync_with_stdio();
cin>>t;
while(t--){
for(int i=;i<=;i++)cin>>x_1[i]>>y_1[i]>>x_2[i]>>y_2[i];
a=,b=;
if(x_1[]>x_1[])a^=,b^=;
if(x_1[]==x_1[]){
if(x_2[b]>x_2[a]||y_2[b]>y_2[a]||y_1[b]<y_1[a])a^=,b^=;
}
if(x_1[a]==x_1[b]&&x_2[a]==x_2[b]&&y_1[a]==y_1[b]&&y_2[a]==y_2[b]){
cout<<<<endl;
}
else{
if(x_2[a]<=x_1[b]){
cout<<<<endl;
}
else{
if(y_2[a]<=y_1[b]||y_1[a]>=y_2[b]){
cout<<<<endl;
}
else{
if(x_1[a]==x_1[b]){ ///l1
if(x_2[a]==x_2[b]){ ///r1
if(y_1[a]==y_1[b]){ ///d1
if(y_2[a]==y_2[b]) ///u1
cout<<<<endl;
else
cout<<<<endl;
}
else{ ///d0
if(y_2[a]==y_2[b]) ///u1
cout<<<<endl;
else ///u0
cout<<<<endl;
}
}
else if(x_2[b]>x_2[a]){ ///l1 r0
if(y_1[a]==y_1[b]){ ///d1
if(y_2[a]==y_2[b]||y_2[a]<y_2[b])
cout<<<<endl;
else if(y_2[a]>y_2[b])
cout<<<<endl;
}
else{ ///d0
if(y_1[a]<y_1[b]){
if(y_2[a]==y_2[b])cout<<<<endl;
else if(y_2[a]>y_2[b]) cout<<<<endl;
else cout<<<<endl;
}
else{
if(y_2[a]==y_2[b])cout<<<<endl;
else if(y_2[a]>y_2[b]) cout<<<<endl;
else cout<<<<endl;
}
}
}
else{
if(y_1[a]==y_1[b]){ ///d1
if(y_2[a]==y_2[b]||y_2[a]>y_2[b])
cout<<<<endl;
else if(y_2[a]<y_2[b])
cout<<<<endl;
}
else{ ///d0
if(y_1[a]>y_1[b]){
if(y_2[a]==y_2[b])cout<<<<endl;
else if(y_2[a]>y_2[b]) cout<<<<endl;
else cout<<<<endl;
}
else{
if(y_2[a]==y_2[b])cout<<<<endl;
else if(y_2[a]>y_2[b]) cout<<<<endl;
else cout<<<<endl;
}
}
}
}
else{
if(x_2[a]==x_2[b]){ ///l0 r1
if(y_1[a]==y_1[b]){
if(y_2[a]==y_2[b])
cout<<<<endl;
else if(y_2[a]>y_2[b])cout<<<<endl;
else cout<<<<endl;
}
else{
if(y_1[a]<y_1[b]){
if(y_2[a]==y_2[b]||y_2[a]>y_2[b])cout<<<<endl;
else cout<<<<endl;
}
else{
if(y_2[a]==y_2[b]||y_2[a]>y_2[b])cout<<<<endl;
else cout<<<<endl;
}
}
}
else{
if(x_2[a]>x_2[b]){
if(y_1[a]==y_1[b]){
if(y_2[a]==y_2[b])cout<<<<endl;
else if(y_2[a]>y_2[b])cout<<<<endl;
else cout<<<<endl;
}
else{
if(y_1[a]<y_1[b]){
if(y_2[a]==y_2[b]||y_2[a]>y_2[b])cout<<<<endl;
else cout<<<<endl;
}
else{
if(y_2[a]==y_2[b])cout<<<<endl;
else if(y_2[a]>y_2[b])cout<<<<endl;
else cout<<<<endl;
}
}
}
else{
if(y_1[a]==y_1[b]){
cout<<<<endl;
}
else{
if(y_1[a]<y_1[b]){
cout<<<<endl;
}
else{
if(y_2[a]==y_2[b]||y_2[a]>y_2[b])cout<<<<endl;
else cout<<<<endl;
}
}
}
}
}
}
}
}
} }

[枚举] HDU 2019 Multi-University Training Contest 8 - Calabash and Landlord的更多相关文章

  1. hdu 4864 Task---2014 Multi-University Training Contest 1

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4864 Task Time Limit: 4000/2000 MS (Java/Others)    M ...

  2. hdu 4937 2014 Multi-University Training Contest 7 1003

    Lucky Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) T ...

  3. hdu 4946 2014 Multi-University Training Contest 8

    Area of Mushroom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) ...

  4. HDU 6395 2018 Multi-University Training Contest 7 (快速幂+分块)

    原题地址 Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)T ...

  5. hdu 4941 2014 Multi-University Training Contest 7 1007

    Magical Forest Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Other ...

  6. hdu 4939 2014 Multi-University Training Contest 7 1005

    Stupid Tower Defense Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/ ...

  7. HDU - 6304(2018 Multi-University Training Contest 1) Chiaki Sequence Revisited(数学+思维)

    http://acm.hdu.edu.cn/showproblem.php?pid=6304 题意 给出一个数列的定义,a[1]=a[2]=1,a[n]=a[n-a[n-1]]+a[n-1-a[n-2 ...

  8. hdu 5755 2016 Multi-University Training Contest 3 Gambler Bo 高斯消元模3同余方程

    http://acm.hdu.edu.cn/showproblem.php?pid=5755 题意:一个N*M的矩阵,改变一个格子,本身+2,四周+1.同时mod 3;问操作多少次,矩阵变为全0.输出 ...

  9. hdu 5738 2016 Multi-University Training Contest 2 Eureka 计数问题(组合数学+STL)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5738 题意:从n(n <= 1000)个点(有重点)中选出m(m > 1)个点(选出的点只 ...

随机推荐

  1. ansible使用指北(二)

    前言在上一篇文章里我们了解了ansible的常用模块,今天我们来了解下ansible-playbook,ansbile-playbook是一系统ansible命令的集合,其利用yaml 语言编写,an ...

  2. 【底层原理:深入理解计算机系统】#1 一切从"hello world"说起 (一)

    计算机系统是由硬件和系统软件组成的,他们共同工作来运行应用程序.虽然系统的具体实现方式随着时间不断的在变化,但是系统的内在概念却没有改变的. 所有的计算机硬件和软件有着相似的结构和功能.这个系列专题便 ...

  3. STL迭代器的使用、正向、逆向输出双向链表中的所有元素

    */ * Copyright (c) 2016,烟台大学计算机与控制工程学院 * All rights reserved. * 文件名:text.cpp * 作者:常轩 * 微信公众号:Worldhe ...

  4. python库常用函数学习

    os.path #返回标准化的绝对路径,基本等同于normpath() os.path.abspath(path) #返回文件名 os.path.basename(path) #返回目录名 os.pa ...

  5. pycharm 关于模块安装出现的“[error] Microsoft Visual C++ 14.0 is required” 解决办法

    刚才正准备对pycharm进行一番操作的时候,噔  噔磴噔噔 “no module define xxx” ,那我当然要把xxx给搞到pycharm上来啊, 不一会功夫 ,biu~ “[error] ...

  6. 工作技术点小计14条 hybrid + animate 方向

    设置transition 动画的时候 , js直接设置duration 和 变化值不会起作用 , 需要先设置duration , 等一小会再设置变化值 安卓端 , 窗口不可见时 , window.in ...

  7. tab 切换下划线跟随实现

    HTML 结构如下: <ul> <li class="active">不可思议的CSS</li> <li>导航栏</li> ...

  8. 2019年最新老男孩高性能Web架构与自动化运维架构视频教程

    课程目录L001-老男孩架构15期-Web架构之单机时代L002-老男孩架构15期-Web架构之集群时代L003-老男孩架构15期-Web架构之dnsL004-老男孩架构15期-Web架构之缓存体系L ...

  9. 深入学习JAVA注解-Annotation(学习过程)

    JAVA注解-Annotation学习 本文目的:项目开发过程中遇到自定义注解,想要弄清楚其原理,但是自己的基础知识不足以支撑自己去探索此问题,所以先记录问题,然后补充基础知识,然后解决其问题.记录此 ...

  10. form里面文件上传并预览

    其实form里面是不能嵌套form的,如果form里面有图片上传和其他input框,我们希望上传图片并预览图片,然后将其他input框填写完毕,再提交整个表单的话,有两种方式! 方式一:点击上传按钮的 ...