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 <…

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…

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…

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…

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: 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…

/** 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…

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…