Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7632   Accepted: 2263   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: 11539   Accepted: 3395   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…

题链: http://poj.org/problem?id=3608 题解: 计算几何,求两个凸包间的最小距离,旋转卡壳 两个凸包间的距离,无非下面三种情况: 所以可以基于旋转卡壳的思想,去求最小距离. (分别用i,j表示A,B凸包上枚举到的点,i的初始位置为A上y最小的顶点,j的初始位置为B上y最大的顶点.) 逆时针枚举凸包A的每一条边$\vec{A_iA_{i+1}}$,然后对另一个凸包B逆时针旋转卡壳,找到第一个$\vec{B_{j+1}B_j}\times\vec{A_iA_{i+1}}…

题目链接:http://poj.org/problem?id=2187 旋转卡壳算法:http://www.cppblog.com/staryjy/archive/2009/11/19/101412.html 或 http://cgm.cs.mcgill.ca/~orm/rotcal.frame.html #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #includ…

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…

给两个凸包,求这两个凸包间最短距离 旋转卡壳的基础题 因为是初学旋转卡壳,所以找了别人的代码进行观摩..然而发现很有意思的现象 比如说这个代码(只截取了关键部分) double solve(Point* P, Point* Q, int n, int m) { , ymaxQ = ; ; i < n; ++i) if (P[i].y < P[yminP].y) yminP = i; // P上y坐标最小的顶点 ; i < m; ++i) if (Q[i].y > Q[ymaxQ].…

The Fortified Forest Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6198   Accepted: 1744 Description Once upon a time, in a faraway land, there lived a king. This king owned a small collection of rare and valuable trees, which had been…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 33888   Accepted: 11544 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he w…

Poj 2187 凸包模板求解 传送门 由于整个点数是50000,而求凸包后的点也不会很多,因此直接套凸包之后两重循环即可求解 #include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #define ll long lo…

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:/…