简单几何(极角排序) POJ 2007 Scrambled Polygon

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 <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int dcmp(double x)  {
    if (fabs (x) < EPS) return 0;
    else    return x < 0 ? -1 : 1;
}
struct Point   {
    double x, y;
    Point ()    {}
    Point (double x, double y) : x (x), y (y) {}
    Point operator - (const Point &r) const {       //向量减法
        return Point (x - r.x, y - r.y);
    }
    bool operator == (const Point &r) const {       //判断同一个点
        return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
    }
    bool operator < (const Point &r) const  {
        return x < r.x || (dcmp (x - r.x) == 0 && y < r.y);
    }
};
typedef Point Vector;
Point read_point(void)  {
    double x, y;
    scanf ("%lf%lf", &x, &y);
    return Point (x, y);
}
double polar_angle(Vector A)    {
    return atan2 (A.y, A.x);
}
double dot(Vector A, Vector B)  {
    return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B)    {
    return A.x * B.y - A.y * B.x;
}

bool cmp(Point a, Point b)  {
    return cross (a - Point (0, 0), b - Point (0, 0)) > 0;
}

int main(void)    {
    vector<Point> ps;
    double x, y;
    while (scanf ("%lf%lf", &x, &y) == 2)   {
        ps.push_back (Point (x, y));
    }
    sort (ps.begin ()+1, ps.end (), cmp);
    for (int i=0; i<ps.size (); ++i)    {
        printf ("(%.0f,%.0f)\n", ps[i].x, ps[i].y);
    }

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

    return 0;
}