凸包(旋转卡壳) 大概理解了凸包A了两道模板题之后在去吃饭的路上想了想什么叫旋转卡壳呢?回来无聊就搜了一下,结果发现其范围真广. 凸包: 凸包就是给定平面图上的一些点集(二维图包),然后求点集组成的凸多边形,但是要包含所有的点. 求凸多边形的方法:Graham算法描述如下: Graham()算法先对点进行排序,有极角序和水平序两种排序方式.我们仍然以左下方的点作为基准点来通过叉积进行排序.利用STL里面的sort或者自己写个排序算法,排序所用时间为O(NlogN),极角序列如左图所示:极角排序

旋转卡壳求凸包直径. 参考:http://www.cppblog.com/staryjy/archive/2010/09/25/101412.html #include <cstdio> #include <cmath> #include <algorithm> using namespace std; << ; struct Point { int x, y; Point( , ):x(x), y(y) { } }; typedef Point Vecto

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

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; struct Point{ int x,y; Point(int x=0,int y=0):x(x),y(y){} }; typedef Point Vector; bool cmp(const Point &p,con

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

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&

#include<iostream> #include<cstdio> #include<cmath> #include<vector> #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 Po

G++一直没有过了 换成 C++果断A掉了...It's time to bet RP. 题意:给一个多边形,然后放进去两个圆,让两个圆的覆盖面积尽量最大,输出两个圆心的坐标. 思路:将多边形的边向里平移圆的的半径R,然后求新多边形的距离最长的两个点. 平移多少废了一点脑筋,其他的就都是现成的模板了. 这个是平移的函数,自己想得,不知道还有没有更简便的.左右平移只需要改一下 向量 V void Panning_Edge(P &a1,P &a2,double dis) { //向v的右侧平移

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 26180   Accepted: 8081 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

1.HDU 1392 Surround the Trees 2.题意:就是求凸包周长 3.总结:第一次做计算几何,没办法,还是看了大牛的博客 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #include<cstdlib> #define F(i,a,b) f

http://poj.org/problem?id=1113 题目大意:现在要给n个点,让你修一个围墙把这些点围起来,距离最小是l 分析  :现在就是求凸包的周长然后再加上一个圆的周长 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<algorithm> #include<iostream> #include<qu

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28157   Accepted: 9401 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 wo

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28462   Accepted: 9498 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 wo

题目传送门 题意:判断一些点的凸包能否唯一确定 分析:如果凸包边上没有其他点,那么边想象成橡皮筋,可以往外拖动,这不是唯一确定的.还有求凸包的点数<=2的情况一定不能确定. /************************************************ * Author :Running_Time * Created Time :2015/11/4 星期三 10:24:45 * File Name :POJ_1228.cpp *************************

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1348 求凸包周长+2*PI*L: #include <stdio.h> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; ; ); struct point { double x, y; point(){} point(double x, double

题目链接 题意 : 求凸包周长+一个完整的圆周长. 因为走一圈,经过拐点时,所形成的扇形的内角和是360度,故一个完整的圆. 思路 : 求出凸包来,然后加上圆的周长 #include <stdio.h> #include <string.h> #include <iostream> #include <cmath> #include <algorithm> const double PI = acos(-1.0) ; using namespac

两次DFS求树直径方法见 这里. 这里的直径是指最长链包含的节点个数,而上一题是指最长链的路径权值之和,注意区分. K <= R: ans = K − 1; K > R:   ans = R − 1 + ( K − R ) ∗ 2; #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; ; struct n

题目大意:给N个点,然后要修建一个围墙把所有的点都包裹起来,但是要求围墙距离所有的点的最小距离是L,求出来围墙的长度. 分析:如果没有最小距离这个条件那么很容易看出来是一个凸包,然后在加上一个最小距离L,那么就是在凸包外延伸长度为L,如下图,很明显可以看出来多出来的长度就是半径为L的圆的周长,所以总长度就是凸包的周长+半径为L的圆的周长. 代码如下: -------------------------------------------------------------------------

Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25256   Accepted: 7756 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

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31199   Accepted: 10521 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

题目链接 大意: 求凸包的面积. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue>