A. Wilbur and Swimming Pool
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the area of the rectangle
must be positive. Wilbur had all four vertices of the planned pool written on a paper, until his friend came along and erased some of the vertices.

Now Wilbur is wondering, if the remaining n vertices of the initial rectangle give enough information to restore the area of the planned
swimming pool.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 4) —
the number of vertices that were not erased by Wilbur's friend.

Each of the following n lines contains two integers xi and yi ( - 1000 ≤ xi, yi ≤ 1000) —the
coordinates of the i-th vertex that remains. Vertices are given in an arbitrary order.

It's guaranteed that these points are distinct vertices of some rectangle, that has positive area and which sides are parallel to the coordinate axes.

Output

Print the area of the initial rectangle if it could be uniquely determined by the points remaining. Otherwise, print  - 1.

Sample test(s)
input
2
0 0
1 1
output
1
input
1
1 1
output
-1
Note

In the first sample, two opposite corners of the initial rectangle are given, and that gives enough information to say that the rectangle is actually a unit square.

In the second sample there is only one vertex left and this is definitely not enough to uniquely define the area.

水题、但是窝过的也很水、想复杂了、题目意思是一个矩形抹去了若干点、前提是这本身就是个矩形,可是窝刚理解为任意四点、、、题意啊!写的还那么复杂、、渣

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std; struct node {
int x, y;
}a[4]; int mianJi(int a1, int b1, int a2, int b2) {
return abs(a1-a2) * abs(b1-b2);
} int main() {
int n;
memset(a, 0, sizeof(a)); cin >> n;
for (int i = 0; i<n; i++)
cin >> a[i].x >> a[i].y;
if (n == 1)
cout << -1 << endl;
else {
if (n == 2) {
if (a[0].x != a[1].x && a[0].y != a[1].y)
cout <<mianJi(a[0].x, a[0].y, a[1].x, a[1].y)<< endl;
else
cout << -1 << endl;
}
else if (n == 3) {
int flag1 = 0;
int flag2 = 0;
for (int i = 0; i<3; i++) {
for (int j = i+1; j<3; j++) {
if (a[i].x == a[j].x)
flag1 = 1;
if (a[i].y == a[j].y)
flag2 = 1;
}
}
if (flag1 && flag2) {
for (int i = 0; i<3; i++) {
for (int j = i+1; j<3; j++) {
if (a[i].x != a[j].x && a[i].y != a[j].y)
cout << mianJi(a[i].x, a[i].y, a[j].x, a[j].y) << endl;
}
}
}
else
cout << -1 << endl;
}
else {
int flag3 = 0;
int flag4 = 0;
for (int i = 0; i<4; i++) {
for (int j = i+1; j<4; j++) {
if (a[i].x == a[j].x)
flag3 ++ ;
if (a[i].y == a[j].y)
flag4 ++ ;
}
}
if (flag3 == 2 && flag4 == 2) {
for (int i = 0; i<3; i++) {
for (int j = i+1; j<3; j++) {
if (a[i].x != a[j].x && a[i].y != a[j].y)
cout << mianJi(a[i].x, a[i].y, a[j].x, a[j].y) << endl;
}
}
}
else
cout << -1 << endl;
}
}
return 0;
}

一下附上经典代码吧、

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std; int main()
{
int n;cin>>n;
int minx,maxx,miny,maxy;
cin>>minx>>miny;
maxx=minx,maxy=miny;
if(n==1)return puts("-1");
for(int i=1;i<n;i++)
{
int x,y;cin>>x>>y;
minx=min(minx,x);
maxx=max(maxx,x);
miny=min(miny,y);
maxy=max(maxy,y);
}
if((maxx==minx)||(maxy==miny))
return puts("-1");
printf("%d\n",(maxx-minx)*(maxy-miny));
}

多想多思考、看到题目不能脑海里浮现出思路就开始码代码,还要思考一下自己的想法是最优的或者可行与否、是否还存在更优的求解方案!??这样才会避免过多的wa!都这么长时间了还停留在div2的AB上也的确要反洗自己了、、、、

Codeforces Round #331 (Div. 2) _A. Wilbur and Swimming Pool的更多相关文章

  1. Codeforces Round #331 (Div. 2) A. Wilbur and Swimming Pool 水题

    A. Wilbur and Swimming Pool Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/conte ...

  2. Codeforce#331 (Div. 2) A. Wilbur and Swimming Pool(谨以此题来纪念我的愚蠢)

    C time limit per test 1 second memory limit per test 256 megabytes input standard input output stand ...

  3. Codeforces Round #331 (Div. 2) E. Wilbur and Strings dfs乱搞

    E. Wilbur and Strings Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/596 ...

  4. Codeforces Round #331 (Div. 2) D. Wilbur and Trees 记忆化搜索

    D. Wilbur and Trees Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/596/p ...

  5. Codeforces Round #331 (Div. 2)C. Wilbur and Points 贪心

    C. Wilbur and Points Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/596/ ...

  6. Codeforces Round #331 (Div. 2) B. Wilbur and Array 水题

    B. Wilbur and Array Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/596/p ...

  7. Codeforces Round #331 (Div. 2) C. Wilbur and Points

    C. Wilbur and Points time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

  8. Codeforces Round #331 (Div. 2) B. Wilbur and Array

    B. Wilbur and Array time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  9. Codeforces Round #331 (Div. 2)

    水 A - Wilbur and Swimming Pool 自从打完北京区域赛,对矩形有种莫名的恐惧.. #include <bits/stdc++.h> using namespace ...

随机推荐

  1. 【Zookeeper】源码分析之服务器(一)

    一.前言 前面已经介绍了Zookeeper中Leader选举的具体流程,接着来学习Zookeeper中的各种服务器. 二.总体框架图 对于服务器,其框架图如下图所示 说明: ZooKeeperServ ...

  2. php iconv 函数参数的区别

    本文同时发表在https://github.com/zhangyachen/zhangyachen.github.io/issues/57 用户输入:英特尔® 酷睿™ i7处理器大显身手 case1 ...

  3. 【WebGL】《WebGL编程指南》读书笔记——第5章

    一.前言        终于到了第五章了,貌似开始越来越复杂了. 二.正文         Example1:使用一个缓冲区去赋值多个顶点数据(包含坐标及点大小) function initVerte ...

  4. KVM(一):KVM安装

    KVM通俗的说就是一台服务器当多台用,详细介绍去百度和谷歌. 首先查看服务器是否支持虚拟化 [root@KVM ~]# grep -E '(vmx|svm)' /proc/cpuinfo --colo ...

  5. pstree 命令详解

    作用: 以命令树状图的方式展现进程之间的派生关系, 显示效果比较直观. 选项: -a 显示每个程序的完整指令, 包含路径, 参数或者是常驻服务的标志 -c 不使用精简标示法 -h 列出树状图,特别标明 ...

  6. hello world 为什么我们看到学习中有这一句话!!!

    Hello World ,中文意思:你好,世界.世界上的第一个程序就是Hello World,由Brian Kernighan创作. Hello, world"程序是指在计算机屏幕上输出“H ...

  7. iOS音频采集过程中的音效实现

    1.背景 在移动直播中, 声音是主播和观众互动的重要途径之一, 为了丰富直播的内容,大家都会想要在声音上做一些文章, 在采集录音的基础上玩一些花样. 比如演唱类的直播间中, 主播伴随着背景音乐演唱. ...

  8. python 爬去拉钩测试招聘信息

    代码如下: #coding:utf-8 import time import urllib.request from bs4 import BeautifulSoup file=open(r'meit ...

  9. input选择框样式修改与自定义

    html自带的选择框样式不好看,并且在ios设备上丑的罚款.所以一般都是自定义样式: 原理:将原来默认的input选择框隐藏,然后控制label的:before与:after,配合矢量图标或者图片来实 ...

  10. 转载:DNS解析过程详解

    2015-09-20 此好文是转载,如有侵权联系我,立马删掉 DNS的几个基本概念: 一. 根域 就是所谓的“.”,其实我们的网址www.baidu.com在配置当中应该是www.baidu.com. ...