题意: 有一个长方形,里面从左到右有n条线段,将矩形分成n+1个格子,编号从左到右为0~n. 端点分别在矩形的上下两条边上,这n条线段互不相交. 现在已知m个点,统计每个格子中点的个数. 分析: 用叉积判断点与线段的相对位置,对于每个点二分查找所在的格子. #include <cstdio> #include <cmath> #include <cstring> struct Point { int x, y; Point(, ):x(x), y(y) {} }; ty…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13262   Accepted: 6412 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…

题目链接:http://poj.org/problem?id=2318 #include<stdio.h> #include<cstdlib> #include<cstring> #include<algorithm> #include<queue> #include <stack> using namespace std; #define ll long long #define INF 0x3f3f3f3f #define met…

用叉积判断左右 快速读入写错了卡了3小时hhh #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> #define N 5003 #define read(x) x = getint() using namespace std; inline int getint() { int fh = 1, k = 0; char c = getchar(); for(; c &l…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13120   Accepted: 6334 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…

TOYS   Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his…

BUPT 2017 summer training (16) #2 A 题意 有一个玩具盒,被n个隔板分开成左到u右n+1个区域,然后给每个玩具的坐标,求每个区域有几个玩具. 题解 依次用叉积判断玩具在每个隔板左边还是右边. 知识 设\(\vec a=(x_1,y_1),\vec b=(x_2,y_2)\). 点乘:内积(数量积),\(\vec{a}\cdot\vec{b}=|a||b|cos\theta=x_1\cdot x_2+y_1\cdot y_2\) 叉乘:外积(向量积),\(|\ve…

2318 2398 题意:给出n条线将一块区域分成n+1块空间,再给出m个点,询问这些点在哪个空间里. 思路:由于只要求相对位置关系,而对具体位置不关心,那么易使用叉积性质得到相对位置关系(左侧/右侧),再因为是简单几何线段不相较,即有序分布,那么在求在哪个区间时可以先对所有线段根据x坐标排序,使用二分减少复杂度. /** @Date : 2017-07-11 11:05:59 * @FileName: POJ 2318 叉积性质.cpp * @Platform: Windows * @Auth…

题目传送门:POJ 2318 TOYS Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular bo…

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…

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…

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 ****…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5016   Accepted: 2978 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…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3967   Accepted: 2489 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19…

http://poj.org/problem?id=2318 TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10178   Accepted: 4880 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8661   Accepted: 4114 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away wh…

题目: Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toy…

今天开始学习计算几何,百度了两篇文章,与君共勉! 计算几何入门题推荐 计算几何基础知识 题意:有一个盒子,被n块木板分成n+1个区域,每个木板从左到右出现,并且不交叉. 有m个玩具(可以看成点)放在这个盒子里,问每个区域分别有多少个玩具. 思路:首先,用叉积判断玩具是否在木板的左边,再用二分找到符合的最右边的木板,该木板答案加一. #include<stdio.h> #include<string.h> struct point{ int x,y; point(){} point(…

题目大意:给你一个矩形的左上角和右下角的坐标,然后这个矩形有 N 个隔板分割成 N+1 个区域,下面有 M 组坐标,求出来每个区域包含的坐标数.   分析:做的第一道计算几何题目....使用叉积判断方向,然后使用二分查询找到点所在的区域.   代码如下: ==========================================================================================================================…

题目大意:给出一个长方形盒子的左上点,右下点坐标.给出n个隔板的坐标,和m个玩具的坐标,求每个区间内有多少个玩具. 题目思路:利用叉积判断玩具在隔板的左方或右方,并用二分优化查找过程. #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<stdio.h> #include<stdlib.h> #include<que…

题目链接 题意: 给定一个矩形,n个线段将矩形分成n+1个区间,m个点,问这些点的分布. 题解: 思路就是叉积加二分,利用叉积判断点与直线的距离,二分搜索区间. 代码: 最近整理了STL的一些模板,发现真是好用啊orz,为啥以前没发现呢,可能是比较懒吧-.- #include <stdio.h> #include <string.h> #include <cmath> #include <iostream> #include <queue> #i…

题意:给定n(<=5000)条线段,把一个矩阵分成了n+1分了,有m个玩具,放在为位置是(x,y).现在要问第几个位置上有多少个玩具. 思路:叉积,线段p1p2,记玩具为p0,那么如果(p1p2 ^ p1p0) (记得不能搞反顺序,不同的),如果他们的叉积是小于0,那么就是在线段的左边,否则右边.所以,可以用二分找,如果在mid的左边,end=mid-1 否则begin=mid+1.结束的begin,就是第一条在点右边的线段 #include <cstdio> #include <…

题目链接:http://poj.org/problem?id=2318 Time Limit: 2000MS Memory Limit: 65536K Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is fin…

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> #…

// 题意:问你每个区域有多少个点 // 思路:数据小可以直接暴力 // 也可以二分区间 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <map> #include <set> #include <queue> #include <stdlib.h> #include <cmat…

TOYS Time Limit: 2000MS Memory Limit: 65536K Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They g…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9350   Accepted: 4451 Description Calculate the number of toys that land in each bin of a partitioned toy box.  Mom and dad have a problem - their child John never puts his toys away w…

题目大意:有一个矩形盒子,盒子里会有一些木块线段,并且这些线段是按照顺序给出的,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,问最后每个区域有多少个玩具 解题思路:因为线段是有序给出,所以不用排序,判断某个点在哪个区域,采用二分法,将某个点和线段的叉积来判断这个点是在线的左边或者右边,根据这个来二分找出区域 代码和思路参考与此链接:http://blog.csdn.net/wangjian8006 这也算我入门计算几何的第一道题了,首先看了一些有关资料,了解到叉…

TOYS 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. 思路:这道题很水,只是要知道会使用叉乘来表示点在线的上面还是下面: 当a.Xmult(b,c) < 0时,表示在线的上面.之后就是二分的时候,不能直接使用mid来ans[mid]++; 因为只是确定点在这条线的两边,到底是哪一边,具体还要用tmp来判断;(模板题) #include<iostream> #include<cstdio> #includ…

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…