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…

裸的极角排序,但是要把0,0放在第一个(话说这题题目真是巨长,废话也多...) #include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cassert> #include<iomanip> #include<cst…

http://poj.org/problem?id=2007 #include<stdio.h> #include<math.h> #include<algorithm> #include<iostream> #include<string.h> #include<stdlib.h> #include<ctype.h> #include<vector> using namespace std; #define…

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…

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…

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<…