题目链接: Finding Hotels Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 102400/102400 K (Java/Others) Problem Description There are N hotels all over the world. Each hotel has a location and a price. M guests want to find a hotel with an accept…

原题链接 Description 给出个二维平面上的点,每个点有权值.次询问,求所有权值小于等于的点中,距离坐标的欧几里得距离最小的点.如果有多个满足条件的点,输出最靠前的一个. Solution 拿k-d树搞一搞就好啦. 如果一个子树代表的区域中所有点的权值都大于c,或者到所有点的距离都大于当前答案,就跳过不做. Code //Finding Hotels #include <cstdio> #include <cstring> #include <algorithm>…

Finding Hotels http://acm.hdu.edu.cn/showproblem.php?pid=5992 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)Total Submission(s): 2180    Accepted Submission(s): 688 Problem Description There are N hotels all ove…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=5992 Problem Description There are N hotels all over the world. Each hotel has a location and a price. M guests want to find a hotel with an acceptable price and a minimum distance from their locations.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5992 题目链接:https://www.nowcoder.com/acm/contest/207/K Problem DescriptionThere are N hotels all over the world. Each hotel has a location and a price. M guests want to find a hotel with an acceptable pric…

Problem Description There are N hotels all over the world. Each hotel has a location and a price. M guests want to find a hotel with an acceptable price and a minimum distance from their locations. The distances are measured in Euclidean metric. Inpu…

题意:n家旅店,每个旅店都有坐标x,y,每晚价钱z,m个客人,坐标x,y,钱c,问你每个客人最近且能住进去(非花最少钱)的旅店,一样近的选排名靠前的. 思路:KD树模板题 代码: #include<set> #include<map> #include<stack> #include<cmath> #include<queue> #include<vector> #include<cstdio> #include<c…

题意 给出n个酒店的坐标和价格,然后m个查询,每个查询给出一个人的坐标和能承受的最大价格,然后找出在他价格承受范围以内,距离他最近的宾馆,如果有多个,那么输出第一个 分析 kd树的模板题 #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <cmath> using namespace std; +; ; typedef lo…

比较裸的kd-tree,但是比较考验剪枝. 貌似除了经典的矩形距离剪枝之外, 还必须加个剪枝是某个矩形内的最小价格如果大于价格限制的话,则剪枝. #include<cstdio> #include<cmath> #include<algorithm> using namespace std; #define N 200010 #define EPS 0.00000001 #define INF 1000000000000000007.0 #define KD 2 str…

题面 题意:二维平面上有很多点,每个点有个权值,现在给你一个点(很多组),权值v,让你找到权值小于等于v的点中离这个点最近的,相同的输出id小的 题解:很裸的KDtree,但是查询的时候有2个小限制, 1个是要小于等于v,1个是输出最小id 第一个,对每个点判断dis的时候 如果价钱高于v 距离就变为INF 低于v就没有影响 第二个,如果disl或者disr 和当前最近的dis 相等 就继续询问以得到更小的id #include<bits/stdc++.h> typedef long long…