搜索的应用-计算最优解 题目: You are given nn packages of wiwi kg from a belt conveyor in order (i=0,1,...n−1i=0,1,...n−1). You should load all packages onto kk trucks which have the common maximum load PP. Each truck can load consecutive packages (more than or e…

Line belt Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=3400 Mean: 给出两条平行的线段AB, CD,然后一个人在线段AB的A点出发,走向D点,其中,人在线段AB上的速度为P, 在线段CD上的速度为Q,在其他地方的速度为R,求人从A点到D点的最短时间. analyse: 经典的三分套三分. 首先在AB线段上三分,确定一个点,然后再在CD上三分,确定第二个点,计算出answer.也就是嵌套的三分搜索. Ti…

Line belt Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3531    Accepted Submission(s): 1364 Problem Description In a two-dimensional plane there are two line belts, there are two segments AB…

@theboysmithy did a great piece on coming up with an alternate view for a timeline for an FT piece. Here’s an excerpt (read the whole piece, though, it’s worth it): Here is an example from a story recently featured in the FT: emerging- market populat…

Problem Description In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane. How long must he take to travel from A to D?…

题目传送门 题意:定义$Tour \, Belt$为某张图上的一个满足以下条件的点集:①点集中至少有$2$个点②任意两点互相连通③图上两个端点都在这个点集中的边的权值的最小值严格大于图上只有一个端点在这个点集中的边的权值的最大值.现在给你一张$N$个点,$M$条边的图,请给出这张图上所有$Tour\,Belt$中包含的点数的和.$N \leq 5000 , M \leq \frac{N(N - 1)}{2}$ 虽然这道题没有必要用$Kruskal$重构树来写,但是考%你赛的时候写$Kruskal…

D - Conveyor Belts 思路:分块dp, 对于修改将对应的块再dp一次. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define y1 skldjfskldjg #define y2 skldfjsklejg using namespace std; ;…

题目链接 Line belt Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2862    Accepted Submission(s): 1099 Problem Description In a two-dimensional plane there are two line belts, there are two segment…

HDU 3400 Line belt (三分再三分) ACM 题目地址:  pid=3400" target="_blank" style="color:rgb(0,136,204); text-decoration:none">HDU 3400 Line belt 题意:  就是给你两条线段AB , CD .一个人在AB以速度p跑,在CD上以q跑,在其它地方跑速度是r.问你从A到D最少的时间. 分析:  先三分AB上的点.再三分CD上的点就可以. …

从A出发到D,必定有从AB某个点E出发,从某个点F进入CD 故有E,F两个不确定的值. 在AB上行走的时间   f = AE / p 在其他区域行走的时间 g = EF / r 在CD上行走的时间   h = FD / q 总时间 T = f + g + h 当E确定时,T1 = g + h + C   此时g时一个先减后增的凹函数,h是一个单调递减的凹函数,根据凹函数的性质,故T1是一个凹函数 反之亦然,故需要三分确定其中一个点的位置,再三分另一个点的位置. #include<stdio.h>…