BZOJ 1185([HNOI2007]最小矩形覆盖-旋转卡壳+点集几何意义)

1185: [HNOI2007]最小矩形覆盖

Time Limit: 10 Sec  Memory Limit: 162 MBSec  Special Judge
Submit: 258  Solved: 137

Description

 

l要事先改成r,注意把向量的2点设成同一点会出现奇妙的事情

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
using namespace std;
#define MAXN (50000+10)
#define INF (1000000000)
#define eps 1e-6
struct P
{
    double x,y;
    P(){}
    P(double _x,double _y):x(_x),y(_y){}
    friend bool operator<(P a,P b){return (fabs(a.y-b.y)<eps)?a.x<b.x:a.y<b.y;  }
    friend bool operator==(P a,P b){return fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps;}
    friend bool operator!=(P a,P b){return !(a==b);}

}a[MAXN],s[MAXN],ansp[5];
int size=0;
double ans=INF;
double dis2(P a,P b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}
struct V
{
    double x,y;
    V(){}
    V(double _x,double _y):x(_x),y(_y){}
    V(P a,P b):x(b.x-a.x),y(b.y-a.y){}
    friend double operator*(V a,V b){return a.x*b.y-a.y*b.x;}
    friend V operator*(double a,V b){return V(a*b.x,a*b.y);}
    friend double operator/(V a,V b){return a.x*b.x+a.y*b.y;}
    friend P operator+(P a,V b){return P(a.x+b.x,a.y+b.y);}
    friend P operator-(P a,V b){return P(a.x-b.x,a.y-b.y);}
    friend V operator~(V a){return V(a.y,-a.x);}
    double dis2(){return x*x+y*y;   }   
}c[MAXN];
int cmp(P A,P B)
{
    double tmp=V(a[1],A)*V(a[1],B);
    if (tmp>0) return 1;
    else if (fabs(tmp)<eps) return (-dis2(A,a[1])-dis2(B,a[1])>0);
    return 0;
}
int n;

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lf%lf",&a[i].x,&a[i].y);
    for (int i=2;i<=n;i++) if (a[i]<a[1]) swap(a[1],a[i]);
    sort(a+2,a+1+n,cmp);
    s[1]=a[1];size=1;
    for (int i=2;i<=n;)
        if (size<2||V(s[size-1],s[size])*V(s[size],a[i])>eps) s[++size]=a[i++];
        else size--;
    s[0]=s[size];

    int l=1,r=1,t=1;
    for (int i=0;i<size;i++)
    {
        while (V(s[i],s[i+1])*V(s[i],s[t+1])-V(s[i],s[i+1])*V(s[i],s[t])>-eps) t=(t+1)%size; 
        while (V(s[i],s[i+1])/V(s[i],s[r+1])-V(s[i],s[i+1])/V(s[i],s[r])>-eps) r=(r+1)%size; 
        if (i==0) l=r;
        while (V(s[i],s[i+1])/V(s[i],s[l+1])-V(s[i],s[i+1])/V(s[i],s[l])<eps) l=(l+1)%size; 
        double Dis2=dis2(s[i],s[i+1]),wlxdis=V(s[i],s[i+1])/V(s[i],s[l]),wrxdis=V(s[i],s[i+1])/V(s[i],s[r]),hxdis=V(s[i],s[i+1])*V(s[i],s[t]);
        double tmp=hxdis*(wrxdis-wlxdis)/Dis2;      
        if (tmp<0) tmp=-tmp;
        if (ans>tmp)
        {
            ans=tmp;
            ansp[0]=s[i]-(wlxdis/Dis2)*V(s[i+1],s[i]);
            ansp[1]=s[i]+(wrxdis/Dis2)*V(s[i],s[i+1]);
            ansp[2]=ansp[1]+(hxdis/Dis2)*(~V(s[i+1],s[i]));         
            ansp[3]=ansp[0]+(hxdis/Dis2)*(~V(s[i+1],s[i]));         
        }
    }
    int p=0;
    for (int i=1;i<4;i++) if (ansp[i]<ansp[p]) p=i;//p=0;
    printf("%.5lf\n",ans);    
    for (int i=0;i<4;i++) 
        printf("%.5lf %.5lf\n",ansp[(p+i)%4].x,ansp[(p+i)%4].y);
    return 0;
}