B. The Meeting Place Cannot Be Changed The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost build…

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…

开始推导用公式求了好久(真的蠢),发现精度有点不够. 其实这种凸线上求点类的应该上三分法的,当作入门吧... /** @Date : 2017-09-23 21:15:57 * @FileName: HDU 5144 三分 无聊物理题.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <bit…

Problem Description In the Dark forest, there is a Fairy kingdom where all the spirits will go together and Celebrate the harvest every year. But there is one thing you may not know that they hate walking so much that they would prefer to stay at hom…

题意:给定x轴上有n个点,每一个点都有一个权值,让在x轴上选一个点,求出各点到这个点的距离的三次方乘以权值最小. 析:首先一开始我根本不会三分,也并没有看出来这是一个三分的题目的,学长说这是一个三分的题,我就百度了一下什么是三分算法,一看感觉和二分差不多,当然就是和二分差不多,也是慢慢缩短范围. 这个题也这样,在最左端和最右端不断的三分,直到逼进那个点,刚开始我设置的误差eps是10负8,但是TLE了,我以为是太小,三分数太多,然后我又改成10负6还是TLE,我又失望了,干脆我不用误差了,我让它…

Party All the Time Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4282    Accepted Submission(s): 1355 Problem Description In the Dark forest, there is a Fairy kingdom where all the spirits wil…

题意:n个人,都要去參加活动,每一个人都有所在位置xi和Wi,每一个人没走S km,就会产生S^3*Wi的"不舒适度",求在何位置举办活动才干使全部人的"不舒适度"之和最小,并求最小值. 思路:首先能够得出最后距离之和的表达式最多仅仅有两个极点, 更进一步仅仅有一个极点,否则无最小值. 那么我们就可用三分法或者二分法求解.即对原函数三分或对导数二分就可以. #include<cstdio> #include<cstring> #inclu…

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6979 Accepted Submission(s): 2181 Problem Description In the Dark forest, there is a Fairy kingdom where all the spirits will go together and Celebr…

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…

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…