POJ 3608 Bridge Across Islands

题意

旋转卡壳。

先找第一个凸包上纵坐标最小的点\(p\)和第二个凸包上纵坐标最大的点\(q\)，之后旋转卡壳，求两条线段之间的最短距离。

code:

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=10010;
const double eps=1e-10;
const double inf=0x3f3f3f3f;
int n,m;
struct Point
{
    double x,y;
    inline double len(){return sqrt(x*x+y*y);}
    Point operator+(const Point a)const
    {
        Point res;
        res.x=x+a.x,res.y=y+a.y;
        return res;
    }
    Point operator-(const Point a)const
    {
        Point res;
        res.x=x-a.x,res.y=y-a.y;
        return res; 
    }
    Point operator*(double k)const
    {
        Point res;
        res.x=x*k,res.y=y*k;
        return res; 
    }
    Point operator/(double k)const
    {
        Point res;
        res.x=x/k,res.y=y/k;
        return res; 
    }
    double operator*(const Point a)const{return x*a.y-y*a.x;}
    double operator&(const Point a)const{return x*a.x+y*a.y;}
};
Point st;
Point p1[maxn],p2[maxn];
inline int dcmp(double x)
{
    if(fabs(x)<=eps)return 0;
    return x<0?-1:1;
}
inline Point get(Point a,Point b){return b-a;}
inline double dis1(Point a,Point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
inline double dis2(Point a,Point b,Point c)//a->bc
{
    if(!dcmp(get(b,c).len()))return dis1(a,b);
    if(dcmp(get(b,a)&get(b,c))<0)return dis1(a,b);
    if(dcmp(get(c,a)&get(c,b))<0)return dis1(a,c);
    return fabs((get(a,b)*get(a,c))/dis1(b,c));
}
inline double dis3(Point a,Point b,Point c,Point d)
{
    return min(min(dis2(a,c,d),dis2(b,c,d)),min(dis2(c,a,b),dis2(d,a,b)));
}
inline double solve(Point* p1,Point* p2,int n,int m)
{
    int p=0,q=0;
    for(int i=0;i<n;i++)if(p1[i].y<p1[p].y)p=i;
    for(int i=0;i<m;i++)if(p2[i].y>p2[q].y)q=i;
    p1[n]=p1[0];p2[m]=p2[0];
    double res=inf;
    for(int i=0;i<n;i++)
    {
        while(dcmp(get(p1[p],p1[p+1])*get(p2[q],p2[q+1]))>0)q=(q+1)%m;
        res=min(res,dis3(p1[p],p1[p+1],p2[q],p2[q+1]));
        p=(p+1)%n;
    }
    return res;
}
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    while(~scanf("%d%d",&n,&m)&&n&&m)
    {
        for(int i=0;i<n;i++)scanf("%lf%lf",&p1[i].x,&p1[i].y);
        for(int i=0;i<m;i++)scanf("%lf%lf",&p2[i].x,&p2[i].y);
        for(int i=0;i<n-2;i++)
            if(dcmp(get(p1[i],p1[i+1])*get(p1[i+1],p1[i+2]))<0){reverse(p1,p1+n);break;}
            else if(dcmp(get(p1[i],p1[i+1])*get(p1[i+1],p1[i+2]))>0)break;  
        for(int i=0;i<m-2;i++)
            if(dcmp(get(p2[i],p2[i+1])*get(p2[i+1],p2[i+2]))<0){reverse(p2,p2+m);break;}
            else if(dcmp(get(p2[i],p2[i+1])*get(p2[i+1],p2[i+2]))>0)break;
        printf("%.5lf\n",solve(p1,p2,n,m));
    }
    return 0;
}

来源：https://www.cnblogs.com/nofind/p/12204140.html