CodeForces 772B Volatile Kite

Posted 2020-09-17 Fighting Heart

计算几何，直觉。

凭直觉猜的做法，把每条线段的中点连起来，每个点到对应内部线段的距离，取个最小值。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;

#define eps 1e-8 
#define zero(x)(((x)>0?(x):-(x))<eps)

int T,n;

struct point
{
    double x,y;
};

double xmult(point p1,point p2,point p0)
{ 
    return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); 
}

double distance(point p1,point p2)
{ 
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}

point intersection(point u1,point u2,point v1,point v2)
{ 
    point ret=u1; 
    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) 
            /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
    ret.x+=(u2.x-u1.x)*t;        
    ret.y+=(u2.y-u1.y)*t;
    return ret;
}

point ptoseg(point p,point l1,point l2)
{ 
    point t=p; t.x+=l1.y-l2.y,t.y+=l2.x-l1.x; 
    if(xmult(l1,t,p)*xmult(l2,t,p)>eps) 
        return distance(p,l1)<distance(p,l2)?l1:l2;
    return intersection(p,t,l1,l2);
}

point p[2000];

int main()
{
    
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);

    double ans = 99999999999999.0;
    for(int i=1;i<=n;i++)
    {
        int A = i-1;
        int B = i;
        int C = i+1;

        if(i-1==0) A=n;
        if(i+1==n+1) C=1;

        point a,b;

        a.x = (p[A].x+p[B].x)/2;
        a.y = (p[A].y+p[B].y)/2;

        b.x = (p[C].x+p[B].x)/2;
        b.y = (p[C].y+p[B].y)/2;

        point c = ptoseg(p[B],a,b);
        ans = min(ans,distance(p[B],c));

    }

    printf("%f\n",ans);
    
    return 0;
}