介绍   凸包(Convex Hull)是一个计算几何(图形学)中的概念,它的严格的数学定义为:在一个向量空间V中,对于给定集合X,所有包含X的凸集的交集S被称为X的凸包.   在图像处理过程中,我们常常需要寻找图像中包围某个物体的凸包.凸包跟多边形逼近很像,只不过它是包围物体最外层的一个凸集,这个凸集是所有能包围这个物体的凸集的交集.如下图所示: 在上图中,绿色线条所包围的凸集即为白色图形的凸包.   在opencv中,通过函数convexHulll能很容易的得到一系列点的凸包,比如由点组成的

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

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

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>

B. Polygons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You've got another geometrical task. You are given two non-degenerate polygons A and B as vertex coordinates. Polygon A is stric

1.首先,凸包是啥: 若是在二维平面上,则一般的,给定二维平面上的点集,凸包就是将最外层的点连接起来构成的凸多边型,它能包含点集中所有的点. ─────────────────────────────────────────────────────────────────────────────────────────────────────────── 2.那么,如何通过某种算法求二维平面上的凸包呢? 有Graham扫描法(Graham scan algorithm),复杂度O(nlogn).

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&

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6199   Accepted: 2822 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6812    Accepted Submission(s): 2594 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to

Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2848    Accepted Submission(s): 811 Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a w

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

题意:在某块平面土地上有N个点,你可以选择其中的任意四个点,将这片土地围起来,当然,你希望这四个点围成. 的多边形面积最大.n<=2000. 先求凸包,再枚举对角线,随着对角线的斜率上升,另外两个点的在凸包上的位置也是单调的. 水平扫描法:先将所有点按x排序,然后从左往右边扫边求出上凸壳,然后从右往左扫出下凸壳.最后会发现a[tot]=a[1]. #include<cstdio> #include<algorithm> #define rep(i,l,r) for (int