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…

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}}…

链接: http://poj.org/problem?id=2187 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/E Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24254   Accepted: 7403 Description Bessie, Farmer John's prize cow, h…

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…

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…

[题目链接] http://poj.org/problem?id=3608 [题目大意] 求出两个凸包之间的最短距离 [题解] 我们先找到一个凸包的上顶点和一个凸包的下定点,以这两个点为起点向下一个点画线, 做旋转卡壳,答案一定包含在这个过程中 [代码] #include <cstdio> #include <algorithm> #include <cmath> #include <vector> using namespace std; double E…

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:/…

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].…

/** 旋转卡壳,, **/ #include <iostream> #include <algorithm> #include <cmath> #include <cstdio> using namespace std; ; struct point { double x,y; point(,):x(x),y(y){} }; int dcmp(double x){ if(fabs(x)<eps) ; else ?-:; } point p[],p1[…

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…