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…

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…

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): 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): 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): 612    Accepted Submission(s): 250 Problem Description There are N points in total. Every point moves in certain direction and…

Codeforce 1311 F. Moving Points 解析(思維.離散化.BIT.前綴和) 今天我們來看看CF1311F 題目連結 題目 略,請直接看原題. 前言 最近寫1900的題目更容易不看題解了,不知道是不是較少人\(AC\)的同難度題目會比較簡單. @copyright petjelinux 版權所有 觀看更多正版原始文章請至petjelinux的blog 想法 首先注意到,如果\(x\)座標上的前後兩點\(x_i,x_j\),\(x_i<x_j\),如果\(v[x_i]>v…

Description There are N points in total. Every point moves in certain direction and certain speed. We want to know at what time that the largest distance between any two points would be minimum. And also, we require you to calculate that minimum dist…

这道题看了大家都是用三分做的,其实这道题也是可以用二分来做的,就是利用一下他们的单调性. 对于N个点,总共要考虑N(N+1)/2个距离,距离可以用二次函数表示,而且开口都是向上的. 下面具体说一下二分的过程: 令mid=(L+R)/2,求出在mid时刻的最大距离,同时标记这个最大距离所在的二次函数, 这时候需要判断下mid时刻与对称轴之间的位置关系 1.当mid在对称轴右边时,由于开口是向上的,则最大距离往右是递增的,不可能取到更小值,所以令R=mid: 2.同理,当mid在对称轴左边时,由于开…

题意:坐标系上有n个点,每个点的坐标和移动方向速度告诉你,速度方向都是固定的.然后要求一个时刻,使得这个时刻,这些点中最远的距离最小. 做法:三分法,比赛的时候想不到.考虑两个点,如果它们走出来的路径能在一定时间后相交的话,那么它们之间的距离肯定是先减小后增大,这样其实可以写成一个二次函数(开口朝下),然后考虑所有的点对之间的最远点,就是对所有的二次函数取一个最大值,嗯,好像还是个二次函数,呃,乱想的,想不下去了. 比赛时候想到可能是单调的或者单峰的,现在还是有点想不通,求解答. #define…