CodeForces 800B Volatile Kite(点与直线的距离)(Java 实现)

CodeForces 800B Volatile Kite(点与直线的距离)(Java 实现)

传送门

如果想要一个凸多边形不退化为凹多边形，那么任意的相邻的三个点必然最多形成一条直线。因此我们可以求出点i-1和i+1的直线向量，再求点i到这条直线的距离，答案必然是取其中最小的一个值

import java.io.*;
import java.util.*;

public class Main {
	static class Point {
		double x,y;  
	    Point(double xx,double yy)  {  
	        x=xx;  
	        y=yy;
	    }
	    Point(){
	    	x=0;
	    	y=0;
	    }
	    Point subtract(final Point b)  
	    {  
	        return new Point(x-b.x,y-b.y);  
	    }  
	    Point add(final Point b)  
	    {  
	        return new Point(x+b.x,y+b.y);  
	    }  
	    double multiply(final Point b)  
	    {  
	        return x*b.y-b.x*y;  
	    }  
	    double dis(final Point b)  
	    {  
	        return Math.sqrt((x-b.x)*(x-b.x)+(y-b.y)*(y-b.y));  
	    }  
	}
	static final int N = 10005;
    static final int inf= 0x3f3f3f3f;
    static Point a[]=new Point[N];
	public static void main(String[] args) throws IOException {
		Scanner cin = new Scanner(new InputStreamReader(System.in));
		while(cin.hasNext()){
			int n=cin.nextInt();
			for(int i=0;i<n;i++){
				if(a[i]==null) a[i]=new Point();
				a[i].x=cin.nextDouble();
				a[i].y=cin.nextDouble();
			}
			for(int i=n;i<2*n;i++){
				a[i]=a[i-n];
			}
			double ans=Double.MAX_VALUE;  
		    for(int i=0; i<n; i++)  
		    {  
		        double tmp=a[i+2].subtract(a[i]).multiply(a[i+1].subtract(a[i]))/2;
		        if(a[i].dis(a[i+2])==0) continue;  
		        ans=Math.min(ans,tmp/a[i].dis(a[i+2]));  
		    }
		    System.out.printf("%.7f\n",ans);
		}
		cin.close();
	}
}