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 <stdio.h>#include <iostream>#include <string.h>#include <algorithm>#include <utility>#include <vector>#include <map>#include <set>#include <string>#include <stack>#include <queue>#include <math.h>//#include <bits/stdc++.h>#define LL long long#define PII pair<int ,int>#define MP(x, y) make_pair((x),(y))#define fi first#define se second#define PB(x) push_back((x))#define MMG(x) memset((x), -1,sizeof(x))#define MMF(x) memset((x),0,sizeof(x))#define MMI(x) memset((x), INF, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const int N = 1e5+20;const double eps = 1e-8;struct point{	int x, y;	point(){}	point(int _x, int _y){x = _x, y = _y;}	point operator -(const point &b) const	{		return point(x - b.x, y - b.y);	}	int operator *(const point &b) const 	{		return x * b.x + y * b.y;	}	int operator ^(const point &b) const	{		return x * b.y - y * b.x;	}};double xmult(point p1, point p2, point p0)  {      return (p1 - p0) ^ (p2 - p0);  }  double distc(point a, point b){	return sqrt((double)((b - a) * (b - a)));}point p[N];int cmp(point a, point b){	int t = xmult(a, b, p[0]);	if(t == 0)		return distc(a, p[0]) < distc(b, p[0]);	else 		return t > 0;}int main(){	int x, y;	int cnt = 0;	while(~scanf("%d%d", &x, &y))	{		p[cnt++] = point(x, y);		sort(p + 1, p + cnt, cmp);	}	for(int i = 0; i < cnt; i++)		printf("(%d,%d)\n", p[i].x, p[i].y);    return 0;}

POJ 2007 Scrambled Polygon 极角序 水