Scrambled Polygon POJ

Scrambled Polygon

 POJ - 2007

题意：

 

思路：其实就是将（0，0）这个点按照极角排序，其他点对于（0，0）来排序，将排序后输出就行，注意输入不定

// 
// Created by HJYL on 2020/1/17.
//
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e8;
int dcmp(double x)
{
    if(fabs(x)<eps)
        return 0;
    return x<0?-1:1;
}
struct Point{
    int x,y;
    int id;
    Point(){}
    Point(int id,double x,double y):id(id),x(x),y(y){}
}initial;//以当前点来求最外凸包

typedef Point Vector;

Vector operator-(Point A,Point B)
{
    return Vector(-1,A.x-B.x,A.y-B.y);
}
//叉积
double Cross(Vector A,Vector B)
{
    return A.x*B.y-A.y*B.x;
}

double Len(Point A,Point B)
{
    return (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);
}

//极角排序
bool cmp(const Point& A,const Point& B)
{
    double t=Cross(A-initial,B-initial);
    if(t>0)//叉积大于0，B在A的左边，A在外凸包上
        return true;
    else if(t<0)
        return false;
    else
        return Len(A,initial)<Len(B,initial);//极角相同采用最近的点

}

const int maxn=100;
int main()
{
    Point p[maxn];
    int x,y;
    while(~scanf("%d%d",&x,&y))
    {
        int  num=0;
        initial=Point(0,0,0);
        while(~scanf("%d%d",&p[num].x,&p[num].y)){
            num++;
        }
        printf("(0,0)
");
        for(int i=0;i<num;i++)
        {
            sort(p,p+num,cmp);
            printf("(%d,%d)
",p[i].x,p[i].y);
        }
    }
    return 0;
}