凸包算法实现点集合中搜索凸包顶点的功能,可以处理共线情况,可以输出共线点也可以不输出而只输出凸包顶点.经典的Graham Scan算法,点排序使用极角排序方式,并对共线情况做特殊处理.一般算法是将共线的点去掉距离小的,保留最远的,这样处理会导致不能输出凸包边上的点,只能输出顶点.但是有时候需要输出这些边上的点,因此这里我将共线点都保留,并按照顺序排列.共线点排列方式是:非起始边按照从远道近排列,起始边按从近到远排列. 算法原理参见如下网址,讲解很详细: http://blog.sina.cn/d

题目链接:http://codeforces.com/problemset/problem/70/D 本题关键:在log(n)的复杂度内判断点在凸包 或 把点插入凸包 判断:平衡树log(n)内选出点所属于的区域 插入:平衡树log(n)内选出点所属于的区域, 与做一般凸包的时候类似,分别以该点向左右两边进行维护, 一直删除不满足凸包的点,直到所有点满足凸包为止. 水平序: 可以用2个平衡树分别维护上下2个半凸包,具体实现时可以把其中一个半凸包按y轴对称以后,那么2个半凸包的维护就是同一种方法,

Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 8636   Accepted: 4105 Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the

首先处理处理出来哪些边能连--能把羊分成两个偶数部分的,实现是在凸包上枚举极点,极角排序,枚举凸包上点对判断两边羊的个数的奇偶即可,设可以连边为v[i][j]=1 然后设f[i][j]为从i到j个凸包上点的方案数,初始状态是相邻点f[i][i+1]=1,转移是 \[ f[i][j]=\sum_{k=i+1}^{i-1}[v[i][k]\&\&v[k][j]]f[i][k]*f[k][j] \] #include<iostream> #include<cstdio>

首先肯定是构造一个完整的凸包包括所有的点,那么要使得刚好有两个点在外面,满足这个条件的只有三种情况. 1.两个在凸包上但是不连续的两个点. 2.两个在凸包上但是连续的两个点. 3.一个在凸包上,还有一个在这个点去掉后这段新凸包边上的一个点. 如何快速的截取新凸包的点是谁呢,我们可以将整个凸包划分区域,每个点删掉后,只可能在这块区域内选择新的点.那么我们就可以随机在凸包内部选择一个点,我使用的是凸包的重心作为坐标原点o,那么整个凸包移到原点处,然后在这个点的左侧和右侧的三角形区域内才是有可能构成新

题意:如题 用Graham,直接就能得到逆时针的凸包,找到原点输出就行了,赤果果的水题- 代码: /* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: poj2007.cpp * Create Date: 2013-11-14 18:55:37 * Descripton: convex hull */ #include <cstdio> #include <c

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2876   Accepted: 1839 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19

平面上n个红点,m个黑点,问你多少个黑点至少在一个红三角形内. 对红点求凸包后,转化为询问有多少个黑点在凸包内. 点在凸多边形内部判定,选定一个凸包上的点作原点,对凸包三角剖分,将其他的点极角排序之后,使用二分法就可以判定点在哪个剖分出来的三角形的夹角内,然后用叉积即可判定其在凸包内还是外,O(logn): http://www.cnblogs.com/dream-wind/archive/2012/05/23/2514694.html #include<cstdio> #include<

Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10841   Accepted: 5085 Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4970   Accepted: 3100 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3170   Accepted: 2029 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19

水题,根本不用凸包,就是一简单的极角排序. 叉乘<0,逆时针. #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> using namespace std; const int maxn=55; struct point { double x,y; } p[maxn]; double cross(poi

这道题简直太好了,对于计算几何选手需要掌握的一个方法. 首先对于求解四边形面积,我们可以将四边形按对角线划分成两个三角形,显然此时四边形的面积最大最小值就变成了求解里这个对角线最近最远的点对. 对于此类问题有一个技巧,首先我们将点按照x为第一关键词y为第二关键词从小到大排序,然后我们开始取向量,并将所取的所有向量进行极角排序(斜率排序,叉积排序均可).然后我们可以证明,第一条向量的两个端点在之前排序过的点的序列中是相邻的,如果这两个不是相邻的,说明中间存在一点并且满足中间的点的x大于左端点小于右

题意:平面上有n个点,一只蚂蚁从最左下角的点出发,只能往逆时针方向走,走过的路线不能交叉,问最多能经过多少个点. 思路:每次都尽量往最外边走,每选取一个点后对剩余的点进行极角排序.(n个点必定能走完,这是凸包的性质决定的) #include<stdio.h> #include<string.h> #include<math.h> #include<iostream> #include<algorithm> using namespace std;

题目链接:http://poj.org/problem?id=1696 题意:给你n个点,然后我们用一条线把它们连起来,形成螺旋状: 首先找到左下方的一个点作为起点,然后以它为原点进行极角排序,找到极角最小的那个点,如果又多个选距离近的,每次都这样循环找n个点即可; #include <stdio.h> #include <iostream> #include <algorithm> #include <string.h> using namespace s

1285 - Drawing Simple Polygon   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Given set of points in the plane, your task is to draw a polygon using the points. You have to use all the points. To be more specific, each po

题目传送门 题意:裸的对原点的极角排序,凸包貌似不行. /************************************************ * Author :Running_Time * Created Time :2015/11/3 星期二 14:46:47 * File Name :POJ_2007.cpp ************************************************/ #include <cstdio> #include <al

题目链接 题意 : 对输入的点极角排序 思路 : 极角排序方法 #include <iostream> #include <cmath> #include <stdio.h> #include <algorithm> using namespace std; struct point { double x,y; }p[],pp; double cross(point a,point b,point c) { return (a.x-c.x)*(b.y-c.y

Shanghai Hypercomputers, the world's largest computer chip manufacturer, has invented a new class of nanoparticles called Amphiphilic Carbon Molecules (ACMs). ACMs are semiconductors. It means that they can be either conductors or insulators of elect

Space Ant http://poj.org/problem?id=1696 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5371   Accepted: 3343 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-

Problem Description Matt has a company, Always Cook Mushroom (ACM), which produces high-quality mushrooms.  ACM has a large field to grow their mushrooms. The field can be considered as a 1000 * 1000 grid where mushrooms are grown in grid points numb