Poj 2187 旋转卡壳求解 传送门 旋转卡壳,是利用凸包性质来求解凸包最长点对的线性算法,我们逐渐改变每一次方向,然后枚举出这个方向上的踵点对(最远点对),类似于用游标卡尺卡着凸包旋转一周,答案就在这其中的某个方向上. 直接暴力和旋转卡壳速度对比(仅此题) #include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #inc…

水平序 Graham 扫描算法: 计算二维凸包的时候可以用到,Graham 扫描算法有水平序和极角序两种. 极角序算法能一次确定整个凸包, 但是计算极角需要用到三角函数,速度较慢,精度较差,特殊情况较多. 水平序算法需要扫描两次,但排序简单,讨论简单,不易出错. [算法流程] 1.对顶点按x为第一关键字,y为第二关键字进行排序. 2.准备一个空栈,并将前两个点压入栈. 3.对于每一个顶点A,只要栈顶中还至少两个顶点,记栈顶为T,栈中第二个为U. 若UT(向量) * TA(向量) <= 0, 则将…

Triangle Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 9060   Accepted: 2698 Description Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points. Input…

Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9768   Accepted: 2866   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…

\(\color{#0066ff}{题目描述}\) 几千年前,有一个小王国位于太平洋的中部.王国的领土由两个分离的岛屿组成.由于洋流的冲击,两个岛屿的形状都变成了凸多边形.王国的国王想建立一座桥来连接这两个岛屿.为了把成本降到最低,国王要求你,主教,找到两个岛屿边界之间最小的距离. \(\color{#0066ff}{输入格式}\) 输入由几个测试用例组成. 每个测试用两个整数n,m(3≤n,m≤10000)开始 接下来的n行中的每一行都包含一对坐标,用来描述顶点在一个凸多边形中的位置. 下一条…

思路: 旋转卡壳应用 注意点&边  边&边  点&点 三种情况 //By SiriusRen #include <cmath> #include <cstdio> #include <algorithm> using namespace std; ; ; typedef double db; int n,m; struct P{db x,y;P(){}P(db X,db Y){x=X,y=Y;}}p1[N],p2[N]; P operator-(…

题目链接:http://poj.org/problem?id=3608 #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<queue> using namespace std; ; ; const int INF = 0x3f3f3f; ; const double PI = acos(-…

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…

http://poj.org/problem?id=2187 显然直径在凸包上(黑书上有证明).(然后这题让我发现我之前好几次凸包的排序都错了QAQ只排序了x轴.....没有排序y轴.. 然后本题数据水,暴力也能过... (之前一直以为距离是单增的,其实并不是,应该是三角形面积单增...) 考虑旋转卡壳 一篇好的文章:http://www.cnblogs.com/Booble/archive/2011/04/03/2004865.html 首先对踵点就是两条平行线夹紧凸包的两个点(或者3个点或4…

链接:http://poj.org/problem?id=2187 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the wor…

题目链接:http://poj.org/problem?id=2187 Time Limit: 3000MS Memory Limit: 65536K Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of…

题目:http://poj.org/problem?id=2187 学习材料:https://blog.csdn.net/wang_heng199/article/details/74477738 https://www.jianshu.com/p/74c25c0772d6 可以再倒着枚举一遍那样求凸包. 用叉积算面积来旋转卡壳. 注意在面积等于的时候就不要往后走了,不然只有两个点的数据就会死循环. #include<cstdio> #include<cstring> #inclu…

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…

给定点集的最远两点的距离. 先用graham求凸包.旋(xuán)转(zhuàn)卡(qiǎ)壳(ké)求凸包直径. ps:旋转卡壳算法的典型运用 http://blog.csdn.net/hanchengxi/article/details/8639476. #include <cstdio> #include <cmath> #include <algorithm> #define sqr(x) (x)*(x) #define N 50001 using names…

题面 Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farm…

旋转卡壳求凸包的直径的平方 板子题 #include<cstdio> #include<vector> #include<cmath> #include<algorithm> using namespace std; struct Point { int x, y; Point(int x=0, int y=0):x(x),y(y) { } }; typedef Point Vector; Vector operator - (const Point&…

Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 32708   Accepted: 10156 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a…

Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill bet…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24283   Accepted: 7420 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farmers…

D - Beauty Contest Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a r…

思路: 求个凸包 旋转卡壳一下 就求出来最远点对了 注意共线情况 也就是说   凸包如果有一堆点共线保留端点即可 //By SiriusRen #include <cmath> #include <cstdio> #include <algorithm> using namespace std; ; ,ans; struct P{int x,y;P(){}P(int X,int Y){x=X,y=Y;}}p[N],tb[N]; bool cmp(P a,P b){ret…

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:/…

链接:http://poj.org/problem?id=2079 Triangle Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 8173   Accepted: 2423 Description Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices…

id=2187">Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27218   Accepted: 8410 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a…

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…

[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…

Triangle Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 8917   Accepted: 2650 Description Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points. Input…

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].…

Triangle Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 9525   Accepted: 2845 Description Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points. Input…