POJ 3608 旋转卡壳

题意：

求两个凸包的最近距离。

题解：

原来旋转卡壳这么暴力。。我以前一直以为是O(n)的。。。

画画图，用叉积判断下旋转角度就行了~

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>

#define N 22222
#define PI 3.14159265358979
#define EPS 1e-7

using namespace std;

struct PO
{
    double x,y;
}p1[N],p2[N],o;

int n,m,top1,top2,stk1[N],stk2[N];

inline int doublecmp(double x)
{
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}

inline bool cmp(const PO &a,const PO &b)
{
    if(doublecmp(a.x-b.x)==0) return a.y<b.y;
    return a.x<b.x;
}

inline void read()
{
    for(int i=1;i<=n;i++) scanf("%lf%lf",&p1[i].x,&p1[i].y);
    for(int i=1;i<=m;i++) scanf("%lf%lf",&p2[i].x,&p2[i].y);
}

inline PO operator +(PO a,PO b)
{
    PO c;
    c.x=a.x+b.x;
    c.y=a.y+b.y;
    return c;
}

inline PO operator -(PO a,PO b)
{
    PO c;
    c.x=a.x-b.x;
    c.y=a.y-b.y;
    return c;
}

inline double dot(PO &a,PO &b,PO &c)
{
    return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);
}

inline double cross(PO &a,PO &b,PO &c)
{
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}

inline double getlen(PO &a)//向量的模 
{
    return sqrt(a.x*a.x+a.y*a.y);
}

inline PO getty(PO b,PO a)//投影向量 
{
    PO res=a;
    double k=dot(o,a,b)/(getlen(a)*getlen(a));
    res.x*=k; res.y*=k;
    return res;
}

inline double getangle(PO &a,PO &b,PO&c,PO &d)
{
    PO t=c+(a-d);
    return cross(a,b,t);
}

inline PO rotate(PO a,double hd)
{
    PO c;
    c.x=a.x*cos(hd)-a.y*sin(hd);
    c.y=a.x*sin(hd)+a.y*cos(hd);
    return c;
}

inline double getdis(PO &a,PO &b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

inline double getdis_ps(PO &a,PO &b,PO &c)
{
    PO t=c-b;
    t=rotate(t,-PI*0.5);
    PO s=c-a;;
    PO ty=getty(s,t);
    s=a+ty;
    if(doublecmp(dot(s,b,c))<=0) return getlen(ty);
    else return min(getdis(a,b),getdis(a,c));
}

inline void graham(PO *p,int *stk,int &top,int gs)
{
    sort(p+1,p+1+gs,cmp);
    top=-1;
    stk[++top]=1; stk[++top]=2;
    for(int i=3;i<=gs;i++)
    {
        while(top>=1&&doublecmp(cross(p[stk[top-1]],p[stk[top]],p[i]))<=0) top--;
        stk[++top]=i;
    }
    int tmp=top;
    for(int i=gs-1;i>=1;i--)
    {
        while(top>=tmp+1&&doublecmp(cross(p[stk[top-1]],p[stk[top]],p[i]))<=0) top--;
        stk[++top]=i;
    }
}

inline double rotating_calipers()
{
    int s1=0,s2=0;
    for(int i=1;i<top1;i++)
        if(doublecmp(p1[stk1[i]].y-p1[stk1[s1]].y)<0||(doublecmp(p1[stk1[i]].y-p1[stk1[s1]].y)==0)&&doublecmp(p1[stk1[i]].x-p1[stk1[s1]].x)<0) s1=i;
    for(int i=1;i<top2;i++)
        if(doublecmp(p2[stk2[i]].y-p2[stk2[s2]].y)>0||(doublecmp(p2[stk2[i]].y-p2[stk2[s2]].y)==0)&&doublecmp(p2[stk2[i]].x-p2[stk2[s2]].x)>0) s2=i;
    int t1=s1,t2=s2;
    double ans=getdis(p1[stk1[s1]],p2[stk2[s2]]);
    do
    {
        double af=getangle(p1[stk1[s1]],p1[stk1[(s1+1)%top1]],p2[stk2[s2]],p2[stk2[(s2+1)%top2]]);
        if(doublecmp(af)==0)//卡壳到两条边 
        {
            ans=min(ans,getdis_ps(p1[stk1[s1]],p2[stk2[s2]],p2[stk2[(s2+1)%top2]]));
            ans=min(ans,getdis_ps(p1[stk1[(s1+1)%top1]],p2[stk2[s2]],p2[stk2[(s2+1)%top2]]));
            ans=min(ans,getdis_ps(p2[stk2[s2]],p1[stk1[s1]],p1[stk1[(s1+1)%top1]]));
            ans=min(ans,getdis_ps(p2[stk2[(s2+1)%top2]],p1[stk1[s1]],p1[stk1[(s1+1)%top1]]));
            s1=(s1+1)%top1; s2=(s2+1)%top2;
        }
        else if(doublecmp(af)>0)//卡壳到第一个的边，第二个的点 
        {
            ans=min(ans,getdis_ps(p2[stk2[s2]],p1[stk1[s1]],p1[stk1[(s1+1)%top1]]));
            s1=(s1+1)%top1;
        }
        else//卡壳到第二个的边，第一个的点 
        {
            ans=min(ans,getdis_ps(p1[stk1[s1]],p2[stk2[s2]],p2[stk2[(s2+1)%top2]]));
            s2=(s2+1)%top2;
        }
    }while(t1!=s1||t2!=s2);
    return ans;
}

inline void go()
{
    graham(p1,stk1,top1,n);
    graham(p2,stk2,top2,m);
    printf("%.5lf\n",rotating_calipers());
}

int main()
{
    while(scanf("%d%d",&n,&m),n) read(),go();
    return 0;
}

 这题太坑了，凸包写错了都能ac。。

害得我下一道题直接粘的这个凸包，wa了一下午。。