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。

 

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/************************************************ * 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; }