Bridge Across Islands POJ - 3608 （旋转卡壳求两凸包的最小距离）

思路

具体原理我不是很讲得清，思路和求凸包直径大致相同。
留个模板吧。

代码

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;

const double eps = 1e-8;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
struct Point
{
    double x,y;
    Point(double _x = 0.0,double _y = 0.0)
    {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x, y - b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
    void input()
    {
        scanf("%lf%lf",&x,&y);
    }
};
struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e)
    {
        s = _s;
        e = _e;
    }
};
double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}
Point NearestPointToLineSeg(Point P,Line L)
{
    Point result;
    double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
    if(t >= 0 && t <= 1)
    {
        result.x = L.s.x + (L.e.x - L.s.x)*t;
        result.y = L.s.y + (L.e.y - L.s.y)*t;
    }
    else
    {
        if(dist(P,L.s) < dist(P,L.e))
            result = L.s;
        else result = L.e;
    }
    return result;
}
/*
 * 求凸包，Graham算法
 * 点的编号0~n-1
 * 返回凸包结果Stack[0~top-1]为凸包的编号
 */
const int MAXN = 10010;
Point list[MAXN];
int Stack[MAXN],top;
//相对于list[0]的极角排序
bool _cmp(Point p1,Point p2)
{
    double tmp = (p1-list[0])^(p2-list[0]);
    if(sgn(tmp) > 0)return true;
    else if(sgn(tmp) == 0 && sgn(dist(p1,list[0]) - dist(p2,list[0])) <= 0)
        return true;
    else return false;
}
void Graham(int n)
{
    Point p0;
    int k = 0;
    p0 = list[0];
    //找最下边的一个点
    for(int i = 1;i < n;i++)
    {
        if( (p0.y > list[i].y) || (p0.y == list[i].y && p0.x > list[i].x) )
        {
            p0 = list[i];
            k = i;
        }
    }
    swap(list[k],list[0]);
    sort(list+1,list+n,_cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2)
    {
        top = 2;
        Stack[0] = 0;
        Stack[1] = 1;
        return ;
    }
    Stack[0] = 0;
    Stack[1] = 1;
    top = 2;
    for(int i = 2;i < n;i++)
    {
        while(top > 1 && sgn((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <= 0)
            top--;
        Stack[top++] = i;
    }
}
//点p0到线段p1p2的距离
double pointtoseg(Point p0,Point p1,Point p2)
{
    return dist(p0,NearestPointToLineSeg(p0,Line(p1,p2)));
}
//平行线段p0p1和p2p3的距离
double dispallseg(Point p0,Point p1,Point p2,Point p3)
{
    double ans1 = min(pointtoseg(p0,p2,p3),pointtoseg(p1,p2,p3));
    double ans2 = min(pointtoseg(p2,p0,p1),pointtoseg(p3,p0,p1));
    return min(ans1,ans2);
}
//得到向量a1a2和b1b2的位置关系
double Get_angle(Point a1,Point a2,Point b1,Point b2)
{
    return (a2-a1)^(b1-b2);
}
//旋转卡壳，求两个凸包的最小距离
double rotating_calipers(Point p[],int np,Point q[],int nq)
{
    int sp = 0, sq = 0;
    for(int i = 0;i < np;i++)
        if(sgn(p[i].y - p[sp].y) < 0)
            sp = i;
    for(int i = 0;i < nq;i++)
        if(sgn(q[i].y - q[sq].y) > 0)
            sq = i;
    double tmp;
    double ans = 1e99;
    for(int i = 0;i < np;i++)
    {
        while(sgn(tmp = Get_angle(p[sp],p[(sp+1)%np],q[sq],q[(sq+1)%nq])) < 0 )
            sq = (sq + 1)%nq;
        if(sgn(tmp) == 0)
            ans = min(ans,dispallseg(p[sp],p[(sp+1)%np],q[sq],q[(sq+1)%nq]));
        else ans = min(ans,pointtoseg(q[sq],p[sp],p[(sp+1)%np]));
        sp = (sp+1)%np;
    }
    return ans;
}

double solve(Point p[],int n,Point q[],int m)
{
    return min(rotating_calipers(p,n,q,m),rotating_calipers(q,m,p,n));
}
Point p[MAXN],q[MAXN];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        if(n == 0 && m == 0)break;
        for(int i = 0;i < n;i++)
            list[i].input();
        Graham(n);
        n = top;
        for(int i = 0;i < n;i++)
            p[i] = list[Stack[i]];
        for(int i = 0;i < m;i++)
            list[i].input();
        Graham(m);
        m = top;
        for(int i = 0;i < m;i++)
            q[i] = list[Stack[i]];
        printf("%.5f\n",solve(p,n,q,m));
    }
    return 0;
}