poj3608 Bridge Across Islands】的更多相关文章

地址:http://poj.org/problem?id=3608 题目: Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11259   Accepted: 3307   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the midd…
给你俩凸包,问你它们的最短距离. 咋做就不讲了,经典题,网上一片题解. 把凸包上的点逆时针排序.可以取它们的平均点,然后作极角排序. 旋转卡壳其实是个很模板化的东西…… 先初始化分别在凸包P和Q上取哪个点,一般在P上取纵坐标最小的点,在Q上取纵坐标最大的点 for i=1 to n(n是凸包P上的点数) { while(两条边的叉积>0) { 凸包Q上的点++ } 计算当前两条边之间的答案(或者说每条边的2个端点到另一条边的答案) 凸包P上的点++ } #include<cstdio>…
题目链接 POJ-3608 Bridge Across Islands 题意 依次按逆时针方向给出凸包,在两个凸包小岛之间造桥,求最小距离. 题解 旋转卡壳的应用之一:求两凸包的最近距离. 找到凸包 p 的 y 值最小点 yminP 和 q 的 y 值最大点ymaxQ,然后分别做切线如图. 那么\(AC\times AD> AC\times AB\)则说明B还不是离AC最近的点,所以++ymaxQ. 否则用 \(AC\) 和 \(BD\) 两个线段的距离更新最近距离,并且++yminP,即考察P…
Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10455   Accepted: 3093   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…
Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7202   Accepted: 2113   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…
Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11196   Accepted: 3292   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…
http://poj.org/problem?id=3608 (题目链接) 题意 求两凸包间最短距离 Solution 难写难调,旋转卡壳,还真是卡死我了. 先分别选出两凸包最上点和最下点,从这两点开始向逆时针方向旋转卡壳.用叉乘判断是否旋转旋转,具体操作跟求凸包直径差不多. poj discuss蒯下来的数据制造器: #include<algorithm> #include<iostream> #include<cstdlib> #include<cstdio&…
Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory of the kingdom consists two separated islands. Due to the impact of the ocean current, the shapes of both the islands bec…
题意: 给你两个凸包,求其最短距离. 解法: POJ 我真的是弄不懂了,也不说一声点就是按顺时针给出的,不用调整点顺序. 还是说数据水了,没出乱给点或给逆时针点的数据呢..我直接默认顺时针给的点居然A了,但是我把给的点求个逆时针凸包,然后再反转一下时针顺序,又WA了.这其中不知道有什么玄机.. 求凸包最短距离还是用旋转卡壳的方法,这里采用的是网上给出的一种方法: 英文版:        http://cgm.cs.mcgill.ca/~orm/mind2p.html 中文翻译版:  http:/…
题目:计算两个不相交凸多边形间的最小距离. 分析:计算几何.凸包.旋转卡壳.分别求出凸包,利用旋转卡壳求出对踵点对,枚举距离即可. 注意:1.利用向量法判断旋转,而不是计算角度:避免精度问题和TLE. 2.遇到平行线段时,需要计算4组点到线段距离,不然会漏掉对踵点对. #include <algorithm> #include <iostream> #include <cstdlib> #include <cmath> using namespace std…