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…

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…

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 <…

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…

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…

题目链接:POJ 3304 Problem Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common. Input…

POJ 3304  Segments 题意:给定n(n<=100)条线段,问你是否存在这样的一条直线,使得所有线段投影下去后,至少都有一个交点. 思路:对于投影在所求直线上面的相交阴影,我们可以在那里作一条线,那么这条线就和所有线段都至少有一个交点,所以如果有一条直线和所有线段都有交点的话,那么就一定有解. 怎么确定有没直线和所有线段都相交?怎么枚举这样的直线?思路就是固定两个点,这两个点在所有线段上任意取就可以,然后以这两个点作为直线,去判断其他线段即可.为什么呢?因为如果有直线和所有线段都相…

题目链接:https://vjudge.net/problem/POJ-3304 题意:求是否能找到一条直线,使得n条线段在该直线的投影有公共点. 思路: 如果存在这样的直线,那么在公共投影点作直线的垂线,显然该垂线会经过所有直线,那么原题转换为求是否有经过所有线段的直线. 如果存在这样的直线,那么该直线一定能通过平移和旋转之后经过所有线段中的两个端点,那么我们枚举所有两两线段的端点作为直线的两点,然后是判断直线是否经过所有线段.如果线段为p0p1,直线为p2p3,那么相交时满足:(p0p2^p…

原题 给出一个矩形玩具箱和其中隔板的位置,求每个玩具在第几个隔间内(保证没有在线上的玩具) 将玩具按x轴排序,记录当前隔板的编号,每次判断是否需要右移(左移)隔板(因为是有序的,所以移动次数左右不厚超过1),(即判断在该隔板的左或右边,)这样就可以解决了! #include<cstdio> #include<algorithm> #include<cstring> #define N 5050 using namespace std; int n,m,ans[N]; i…