TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14433   Accepted: 6998 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…

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 toys in, but John i…

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…

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…

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…

                                                             TOYS 2页的提交记录终于搞明白了. 题意:一个盒子由n块挡板分成n+1块区域,有m个玩具,每次给出一个玩具的坐标,最后输出各个区域玩具的数量,玩具不会在线上或者盒子外面,挡板也不会相交.输出格式请看样例! 思路:考察叉积的性质,用一个结构体存线段(两个点共4个坐标),然后输入一个玩具坐标就遍历所有的区域找到然后对应区域加1. 这题看起来也不难,数据5000,开始想着用二分…

题目链接:https://vjudge.net/problem/POJ-2318 题意:有n条线将矩形分成n+1块,m个点落在矩形内,求每一块点的个数. 思路: 最近开始肝计算几何,之前的几何题基本处于挂机状态,但听别人说几何题不会太难,所以打算把几何给过了. 先引入叉积的一个重要性质,O为原点: OP^OQ>0 : P在Q的顺时针方向. OP^OQ<0 : P在Q的逆时针方向. OP^OQ=0 : O,P,Q共线. 那么我们就可以利用该性质判断一个点P在直线AB的左侧当且仅当:PA^PB&l…

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…

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…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10281   Accepted: 4924 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 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…

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…

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…

题目链接:https://cn.vjudge.net/problem/POJ-2318 题意 在一个矩形内,给出n-1条线段,把矩形分成n快四边形 问某些点在那个四边形内 思路 二分+判断点与位置关系 提交过程 WA*n x1和x2,y1和y2在复制的时候没分清(哭 WA 可能存在二分问题? AC 代码 #define PI 3.1415926 #include <cmath> #include <cstdio> #include <vector> #include &…

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…

http://poj.org/problem?id=2398 Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3831   Accepted: 2256 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. T…

嘟嘟嘟 题面:先告诉你一个矩形抽屉的坐标,然后\(n\)个隔板将抽屉分成了\(n + 1\)格(格子从\(0\)到\(n - 1\)标号),接下来随机输入\(m\)个玩具的坐标.问最后每一个格子里有多少个玩具. 仔细想想就是一道计算几何入门题. 对于一个玩具\((x_0, y_0)\),我们只要找到在他的左面且离他最近的隔板\(i\),则这个玩具就在第\(i\)格里. 怎么判断隔板\(AB\)(规定\(A\)点是隔板的上端,\(B\)点是隔板下端)是否在玩具的左边?用叉积即可:只要\(\over…

解题关键:计算几何入门题,通过叉积判断. 两个向量的关系: P*Q>0,Q在P的逆时针方向: P*Q<0,Q在P的顺时针方向: P*Q==0,Q与P共线. 实际就是用右手定则判断的. #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #include<iostream> using namespa…

题目:http://poj.org/problem?id=2318 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.(其中这些线段有序且不相交) 解答: 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. 这是第一道计算几何的题目,怎么说呢,对于二分的边界问题还有点搞不清楚,这次是对线段二分,感觉二分真的很有用处. 开阔思维!!!具…

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…

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…

1.poj2318 TOYS 传送:http://poj.org/problem?id=2318 题意:有m个点落在n+1个区域内.问落在每个区域的个数. 分析:二分查找落在哪个区域内.叉积判断点与线段的位置. #include<iostream> #include<cstring> #include<algorithm> using namespace std; ; struct point{ int x,y; point(){ } point(int _x,int…

图论 图论解题报告索引 DFS poj1321 - 棋盘问题 poj1416 - Shredding Company poj2676 - Sudoku poj2488 - A Knight's Journey poj1724 - ROADS(邻接表+DFS) BFS poj3278 - Catch That Cow(空间BFS) poj2251 - Dungeon Master(空间BFS) poj3414 - Pots poj1915 - Knight Moves poj3126 - Prim…

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: 16042   Accepted: 7688 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<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<queue> using namespace std; ; ; const int INF = 0x3f3f3f; ; const double PI = acos(-…

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…

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…

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…