链接:http://acm.hdu.edu.cn/showproblem.php?pid=4717 n个点,给出初始坐标和移动的速度和移动的方向,求在哪一时刻任意两点之间的距离的最大值的最小. 对于最大值的最小问题首先就会想到二分,然而由于两点之间的距离是呈抛物线状,所以三分的方法又浮上脑海,当然二分也可以做,不过思维上更复杂一些 #include<cmath> #include<cstdio> #include<cstring> #include<iostrea…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4717 思路:三分时间求极小值. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const int MAX_N = (300 + 30); const double e…

The Moving Points Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 72    Accepted Submission(s): 18 Problem Description There are N points in total. Every point moves in certain direction and cer…

The Moving Points Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 710    Accepted Submission(s): 290 Problem Description There are N points in total. Every point moves in certain direction and…

The Moving Points Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 878    Accepted Submission(s): 353 Problem Description There are N points in total. Every point moves in certain direction and c…

The Moving Points Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 612    Accepted Submission(s): 250 Problem Description There are N points in total. Every point moves in certain direction and…

The Moving Points Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2122    Accepted Submission(s): 884 Problem Description There are N points in total. Every point moves in certain direction and…

bzoj 3053 HDU 4347 : The Closest M Points  kd树 题目大意:求k维空间内某点的前k近的点. 就是一般的kd树,根据实测发现,kd树的两种建树方式,即按照方差较大的维度分开(建树常数大)或者每一位轮换分割(询问常数大),后者更快也更好些,以后就果断写第二种了. #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using…

The Moving Points Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 964    Accepted Submission(s): 393 Problem Description There are N points in total. Every point moves in certain direction and c…

斜抛从(0,0)到(x,y),问其角度. 首先观察下就知道抛物线上横坐标为x的点与给定的点的距离与角度关系并不是线性的,当角度大于一定值时可能会时距离单调递减,所以先三分求个角度范围,保证其点一定在抛物线下方,这样距离和角度的关系就是单调的了,再二分角度即可. /** @Date : 2017-09-23 23:17:11 * @FileName: HDU 2298 三分.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail…