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…

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…

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…

/** 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…

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 <…

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…

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…

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…

#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)…

链接:http://poj.org/problem?id=1696 Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3077   Accepted: 1965 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an a…

从TLE的暴力枚举 到 13313MS的扫描线  再到 1297MS的简化后的扫描线,简直感觉要爽翻啦.然后满怀欣喜的去HDU交了一下,直接又回到了TLE.....泪流满面 虽说HDU的时限是2000MS 可是数据也忒强了点吧,真心给HDU跪了. 题意:给定平面上的N个点,属性分别标记为0和1,然后找一条直线,直线上的点全部溶解,一侧的1溶解,另一侧的0溶解.求出最多能溶解的点的个数. 思路:最直接的思路就是O(N^3)的暴力枚举,Discuss里面貌似有大牛过了,肯能是我太过暴力了吧,果断Tl…

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;…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2489   Accepted: 1567 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…

Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 8346   Accepted: 2974 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enc…

http://poj.org/problem?id=1696 极角排序是就是字面上的意思   按照极角排序 题目大意:平面上有n个点然后有一只蚂蚁他只能沿着点向左走  求最多能做多少点 分析:  其实还不知道极角排序到底是什么,   但是又好像知道一点   必须一直排序  然后一直找到最左的点就行了 #include<stdio.h> #include<math.h> #include<algorithm> #include<iostream> #inclu…

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<…

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…

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…

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…

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: 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…

#include<iostream> #include<cmath> #include<algorithm> using namespace std; double eps=1e-8; int sgn(double x) { if(fabs(x)<=eps)return 0; if(x<0)return -1; return 1; } struct Point { double x,y; int index; Point (){} Point(double…

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…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5738 [题目大意] 给出平面中一些点,在同一直线的点可以划分为一个集合,问可以组成多少包含元素不少于2的集合. [题解] 最重要的还是处理点重合,和线重复计算的问题,对于每个点,进行极角排序,作为端点,然后统计包含它的每个集合即可.思路非常简单,然而比赛的时候……一脸懵逼地用map,pair,然后转战hash.看起来对于计算几何的敏感度还有待加强. [代码] #include <cstdio>…

题目链接:点击打开链接 题意: 给定二维坐标上的4个点 问: 找一个点使得这个点距离4个点的距离和最小 输出距离和. 思路: 若4个点不是凸4边形.则一定是端点最优. 否则就是2条对角线的交点最优,能够简单证明一下. 对于凸4边形则先极角排序一下. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef do…

在古老的迈瑞城,巍然屹立着 n 块神石.长老们商议,选取 3 块神石围成一个神坛.因为神坛的能量强度与它的面积成反比,因此神坛的面积越小越好.特殊地,如果有两块神石坐标相同,或者三块神石共线,神坛的面积为 0.000. 长老们发现这个问题没有那么简单,于是委托你编程解决这个难题. 输入格式: 输入在第一行给出一个正整数 n(3 ≤n≤5000).随后 n 行,每行有两个整数,分别表示神石的横坐标.纵坐标(-109≤横坐标.纵坐标<109). 输出格式: 在一行中输出神坛的最小面积,四舍五入保留…

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-…