Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7214   Accepted: 3445 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…

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…

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…

LINK 题意:给出一个简单多边形,按极角序输出其坐标. 思路:水题.对任意两点求叉积正负判断相对位置,为0则按长度排序 /** @Date : 2017-07-13 16:46:17 * @FileName: POJ 2007 凸包极角序.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <…

http://poj.org/problem?id=2007 Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6701   Accepted: 3185 Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments ar…

题链: http://poj.org/problem?id=2007 题解: 计算几何,极角排序 按样例来说,应该就是要把凸包上的i点按 第三像限-第四像限-第一像限-第二像限 的顺序输出. 按 叉积 来排序的确可以A掉,但是显然有错呀. 比如这个例子: 0 0 -2 2 -1 -1 1 0 正确答案显然应该是: (0,0) (-2,2) (-1,-1) (1,0) 但是 用叉积 排序后却是这样: (0,0) (1,0) (-2,2) (-1,-1) (要用叉积排序的话,按道理来讲应该把像限分成…

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…

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…

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…

/** 极角排序输出,,, 主要atan2(y,x) 容易失精度,,用 bool cmp(point a,point b){ 5 if(cross(a-tmp,b-tmp)>0) 6 return 1; 7 if(cross(a-tmp,b-tmp)==0) 8 return length(a-tmp)<length(b-tmp); 9 return 0; 10 } **/ #include <iostream> #include <algorithm> #includ…

#include<stdio.h> #include<algorithm> using namespace std; #define Max 60 struct Point { int x,y; } p[Max]; int xmult(Point p1,Point p2,Point p0) { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } int pos; bool cmp(Point p1,Point p2)…

给一些点,这些点都是一个凸包上的顶点,以第一个点为起点顺时针把别的点拍排一下序列. 分析:最简单的极坐标排序了..................... 代码如下: -------------------------------------------------------------------------------------------------------------------------- #include<iostream> #include<algorithm>…

Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 8005   Accepted: 3798 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…

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…

/* poj 2187 Beauty Contest 凸包:寻找每两点之间距离的最大值 这个最大值一定是在凸包的边缘上的! 求凸包的算法: Andrew算法! */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Point{ Point(){} Point(int x, int y){ this->…

Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7805   Accepted: 3712 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…

Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7799   Accepted: 3707 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…

POJ 2007 将所有的点按逆时针输出 import java.io.*; import java.util.*; public class Main { static class Point implements Comparable<Point>{ double x, y; @Override public int compareTo(Point a) { return (int)cross(a); } public double cross(Point a){ return a.x*y…

题意: 给你n个点,这n个点可以构成一个多边形(但是不是按顺序给你的).原点(0,0)为起点,让你按顺序逆序输出所有点 题解: 就是凸包问题的极角排序 用double一直Wa,改了int就可以了 //原点(0,0)为起点,逆序输出多边形的点 #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> #include<queue> #include<…

Technorati Tags: POJ,计算几何,凸包 初学计算几何,引入polygon后的第一个挑战--凸包 此题可用凸包算法做,只要把压入凸包的点从原集合中排除即可,最终形成图形为螺旋线. 关于凸包,具体可见凸包五法:http://blog.csdn.net/bone_ace/article/details/46239187 此处简述我O(nH)的Jarvis法(H: 凸包上点数) 过程较易于理解. //POJ1696 //凸包变形 //AC 2016.10.13 #include "cs…

也是用模板直接套的题目诶 //#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <cstring> #include <cmath> #include <stack> #include <queue> #include <vector>…

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…

Cows Time Limit: 2000MS Memory Limit: 65536K 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 forced to save money on buying fence posts b…

Ultimate Weapon Time Limit: 2000MS   Memory Limit: 131072K Total Submissions: 2430   Accepted: 1173 Description In year 2008 of the Cosmic Calendar, the Aliens send a huge armada towards the Earth seeking after conquest. The humans now depend on thei…

[题目链接] http://poj.org/problem?id=3246 [题目大意] 给出一些点,请删去一个点,使得包围这些点用的线长最短 [题解] 去掉的点肯定是凸包上的点,所以枚举凸包上的点去掉,再计算面积即可. [代码] #include <cstdio> #include <algorithm> #include <cmath> #include <vector> #include <cstring> using namespace…

[题目链接] http://poj.org/problem?id=2079 [题目大意] 给出一些点,求出能组成的最大面积的三角形 [题解] 最大三角形一定位于凸包上,因此我们先求凸包,再在凸包上计算, 因为三角形在枚举了一条固定边之后,图形面积随着另一个点的位置变换先变大后变小 因此我们发现面积递减之后就移动固定边. [代码] #include <cstdio> #include <algorithm> #include <cmath> #include <ve…

[题目链接] http://poj.org/problem?id=1113 [题目大意] 给出一个城堡,要求求出距城堡距离大于L的地方建围墙将城堡围起来求所要围墙的长度 [题解] 画图易得答案为凸包的周长加一个圆的周长. [代码] #include <cstdio> #include <algorithm> #include <cmath> #include <vector> using namespace std; double EPS=1e-10; co…

题目: http://poj.org/problem?id=1113 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/F Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26219   Accepted: 8738 Description Once upon a time there was a greedy King who…

http://poj.org/problem?id=1113 不多说...凸包网上解法很多,这个是用graham的极角排序,也就是算导上的那个解法 其实其他方法随便乱搞都行...我只是测一下模板... struct POINT{ double x,y; POINT(, ):x(_x),y(_y){}; }; POINT p[MAXN],s[MAXN]; double dist(POINT p1,POINT p2){ return(sqrt((p1.x-p2.x) * (p1.x-p2.x) +…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5472   Accepted: 2334 Description Polygon is a game for one player that starts on a polygon with N vertices, like the one in Figure 1, where N=4. Each vertex is labelled with an integer and…