来源poj 2398 Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing…

题目链接 Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4104   Accepted: 2433 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangula…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5439   Accepted: 3234 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3146   Accepted: 1798 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

2318 TOYS 2398 Toy Storage 题意 : 给你n块板的坐标,m个玩具的具体坐标,2318中板是有序的,而2398无序需要自己排序,2318要求输出的是每个区间内的玩具数,而2318要求输出的是有 i 个玩具的区间有几个. 思路 : 两个题基本差不多,只不过2398排一下序,然后再找个数组标记一下就行. 这个题我一开始没想到用二分,判断了点在四边形内但是没写下去,然后看了网上的二分区间,利用叉积判断点在左边还是右边. 2318代码: #include <stdio.h> #…

Toy Storage 题型与2318 TOYS一样,注意要对线段排序,现在模板又更新了~~ #include<iostream> #include<cstdio> #include<cstring> #include<string.h> #include<algorithm> #include<map> #include<queue> #include<vector> #include<cmath>…

Toy Storage Time Limit: 1000MS Memory Limit: 65536K Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is…

题目传送门:poj 2398 Toy Storage 题目大意:一个长方形的箱子,里面有一些隔板,每一个隔板都可以纵切这个箱子.隔板将这个箱子分成了一些隔间.向其中扔一些玩具,每个玩具有一个坐标,求有\(t​\)个玩具的隔间数(对\(t>0​\)都要输出). 题目分析:涉及到计算几何的知识是求点在线的哪一侧.可以利用叉积来做.取点\(A\)到隔板的上端点\(B\)的向量\(\vec{AB}\)叉乘点\(A\)到隔板的下端点\(C\)的向量\(\vec{AC}\).叉积的公式\(\vec a\ti…

poj2398 Toy Storage 链接 poj 题目大意 这道题的大概意思是先输入6个数字:n,m,x1,y1,x2,y2.n代表卡片的数量,卡片竖直(或倾斜)放置在盒内,可把盒子分为n+1块区域,然后分别用0到n表示,m代表玩具的个数,(x1,y1)代表盒子的左上顶点坐标,(x2,y2)代表盒子的右下顶点坐标,接下来的n行,每行都有两个数a,b,(a,y1),(b,y2)分别代表卡片的两个顶点位置,接下来的m行每行两个数从c,d,(c,d),代表玩具放置的坐标,最后让你输出当区域内玩具的…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4588   Accepted: 2718 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

题意:给定一个如上的长方形箱子,中间有n条线段,将其分为n+1个区域,给定m个玩具的坐标,统计每个区域中的玩具个数. 题解:通过斜率判断一个点是否在两条线段之间. /** 通过斜率比较点是否在两线段之间 */ #include"iostream" #include"cstdio" #include"algorithm" #include"cstring" using namespace std; const int N=100…

题目传送门 题意:POJ 2318 有一个长方形,用线段划分若干区域,给若干个点,问每个区域点的分布情况 分析:点和线段的位置判断可以用叉积判断.给的线段是排好序的,但是点是无序的,所以可以用二分优化.用到了叉积 /************************************************ * Author :Running_Time * Created Time :2015/10/23 星期五 11:38:18 * File Name :POJ_2318.cpp ****…

题目大意:和 TOY题意一样,但是需要对隔板从左到右进行排序,要求输出的是升序排列的含有i个玩具的方格数,以及i值. 题目思路:判断叉积,二分遍历 #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<stdio.h> #include<stdlib.h> #include<queue> #include<…

Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing…

题目链接:http://poj.org/problem?id=2398 Time Limit: 1000MS Memory Limit: 65536K Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in…

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing his toys in…

Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing…

二分点所在区域,叉积判断左右 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; const int N=1005; int T,n,m,x1,y1,x2,y2,ans[N],cnt[N]; struct dian { double x,y; dian(double X=0,d…

题目连接: http://poj.org/problem?id=2318     http://poj.org/problem?id=2398 两题类似的题目,2398是2318的升级版. 题目大概是说,有一个矩形的柜子,中间有一些隔板.告诉你每个隔板的坐标,还有一些玩具的坐标,统计玩具在哪个格子里. 这题的思路很简单,如果玩具在某个隔板的左边和右边叉乘的正负是不同的.如图: 图中点P在线段CD的左边,则向量PC和向量PD叉乘结果小于0.反之P在AB的左边,向量PA叉乘PB大于0. 因此利用这个…

题目链接:http://poj.org/problem?id=2398 思路RT,和POJ2318一样,就是需要排序,输出也不一样.手工画一下就明白了.注意叉乘的时候a×b是判断a在b的顺时针还是逆时针侧,>0是顺时针测,<0是逆时针侧,本题对应看成右.左侧,特别注意. /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏┓┃キリキリ♂ mind! ┛┗┛┗┛┃＼○/ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓…

这道题和POJ 2318几乎是一样的. 区别就是输入中坐标不给排序了,=_=|| 输出变成了,有多少个区域中有t个点. #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> using namespace std; struct Point { int x, y; Point(, ):x(x), y(y) {} }; typedef Point Vector; P…

POJ 2318: 题目大意:给定一个盒子的左上角和右下角坐标,然后给n条线,可以将盒子分成n+1个部分,再给m个点,问每个区域内有多少各点 这个题用到关键的一步就是向量的叉积,假设一个点m在 由abcd围成的四边形区域内,那么向量ab, bc, cd, da和点的关系就是,点都在他们的同一侧,我是按照逆时针来算的,所以只需要判断叉积是否小于0就行了.还有一个问题就是这个题要求时间是2s,所以直接找,不用二分也能过,不过超过1s了,最好还是二分来做,二分时间170ms,二分的时候要把最左边的一条…

计算几何终于开坑了... 叉积+二分. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 5050 using namespace std; struct point { int x,y; point (int x,int y):x(x),y(y) {} point () {} friend point operator-(point…

题目链接:https://cn.vjudge.net/contest/276358#problem/B 题目大意:和上一次写叉积的题目一样,就只是线不是按照顺序给的,是乱序的,然后输出的时候是按照有三个点的区域有多少个---这个类型出发的. AC代码: #include<iostream> #include<cstring> #include<string> #include<cmath> #include<algorithm> #include…

接替关键:和上题类似,输出不同,注意输入这道题需要排序. #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #include<iostream> using namespace std; typedef long long ll; struct point{ int x,y; }; int n,m,x1,x2…

链接:poj 2318 题意:有一个矩形盒子,盒子里有一些木块线段.而且这些线段坐标是依照顺序给出的. 有n条线段,把盒子分层了n+1个区域,然后有m个玩具.这m个玩具的坐标是已知的,问最后每一个区域有多少个玩具 分析:从左往右.直到推断玩具是否在线段的逆时针方向为止.这个就须要用到叉积,当然能够用二分查找优化. 叉积:已知向量a(x1,y1),向量b(x2,y2),axb=x1*y2-x2*y1, 若axb>0,a在b的逆时针方向,若axb<0,则a在b的顺时针方向 注:每组数据后要多空一行…

Solidity是一种智能合约高级语言,运行在Ethereum虚拟机(EVM)之上.这里我会讲解一下关键字storage和memory的区别. storage的结构是在合约部署创建时,根据你的合约中状态变量的声明,就固定下来了,并且不能在将来的合约方法调用中改变这个结构.但是,storage中的内容是可以通过交易来改变的.这些交易调用因此将修改合约的状态. memory关键字告诉solidity应当在该函数运行时为变量创建一块空间,使其大小和结构满足函数运行的需要. 首先局部变量默认是stora…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6534   Accepted: 3905 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3146   Accepted: 1798 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

KUANGBIN带你飞 全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一 简单搜索 POJ 1321 棋盘问题    //2019.3.18 POJ 2251 Dungeon Master POJ 3278 Catch That Cow  //4.8 POJ 3279 Fliptile POJ 1426 Find The Multiple  //4.8 POJ 3126 Prime Path POJ 3087 Shuffle…