[hdu5251]矩形面积 旋转卡壳求最小矩形覆盖

　　旋转卡壳求最小矩形覆盖的模板题。

　　因为最小矩形必定与凸包的一条边平行，则枚举凸包的边，通过旋转卡壳的思想去找到其他3个点，构成矩形，求出最小面积即可。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<cstdlib>
#include<cmath>
#include<list>
using namespace std;
#define MAXN 100100
#define eps 1e-9
#define For(i,a,b) for(int i=a;i<=b;i++) 
#define Fore(i,a,b) for(int i=a;i>=b;i--) 
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mkp make_pair
#define pb push_back
#define cr clear()
#define sz size()
#define met(a,b) memset(a,b,sizeof(a))
#define iossy ios::sync_with_stdio(false)
#define fre freopen
#define pi acos(-1.0)
#define inf 1e6+7
#define Vector Point
const int Mod=1e9+7;
typedef unsigned long long ull;
typedef long long ll;
int dcmp(double x){
    if(fabs(x)<=eps) return 0;
    return x<0?-1:1;
}
struct Point{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {}
    bool operator < (const Point &a)const{
        if(x==a.x) return y<a.y;
        return x<a.x;
    }
    Point operator - (const Point &a)const{
        return Point(x-a.x,y-a.y);
    }
    Point operator + (const Point &a)const{
        return Point(x+a.x,y+a.y);
    }
    Point operator * (const double &a)const{
        return Point(x*a,y*a);
    }
    Point operator / (const double &a)const{
        return Point(x/a,y/a);
    }
    void read(){
        scanf("%lf%lf",&x,&y);
    }
    void out(){
        cout<<"debug: "<<x<<" "<<y<<endl;
    }
    bool operator == (const Point &a)const{
        return dcmp(x-a.x)==0 && dcmp(y-a.y)==0;
    }
};
double Dot(Vector a,Vector b) {
    return a.x*b.x+a.y*b.y;
}
double dis(Vector a) {
    return sqrt(Dot(a,a));
}
double Cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
int ConvexHull(Point *p,int n,Point *ch){
    int m=0;
    For(i,0,n-1) {
        while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    int k=m;
    Fore(i,n-2,0){
        while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    if(n>1) m--;
    return m;
}
double ANS(Point *p,int n){
    int L,R=1,U=1;
    double ans=1e9+9;
    p[n]=p[0];
    For(i,0,n-1) {
        while(Cross(p[i+1]-p[i],p[U+1]-p[i])>=Cross(p[i+1]-p[i],p[U]-p[i])) U=(U+1)%n;
        while(Dot(p[i+1]-p[i],p[R+1]-p[i])>Dot(p[i+1]-p[i],p[R]-p[i])) R=(R+1)%n;
        if(!i) L=R;
        while(Dot(p[i+1]-p[i],p[L+1]-p[i])<=Dot(p[i+1]-p[i],p[L]-p[i])) L=(L+1)%n;
        double tmp=fabs(Cross(p[U]-p[i],p[i+1]-p[i]))/dis(p[i+1]-p[i]);
        double cnt1=fabs(Dot(p[L]-p[i],p[i+1]-p[i]))/dis(p[i+1]-p[i]),cnt2=fabs(Dot(p[R]-p[i],p[i+1]-p[i]))/dis(p[i+1]-p[i]);
        ans=min(ans,(cnt2+cnt1)*tmp);
    }
    return ans;
}
int n,m;
Point p[200005];
Point ch[200005];
void solve(){
    scanf("%d",&n);
    int rt=0;
    For(i,0,n-1) {
        p[rt++].read();
        p[rt++].read();
        p[rt++].read();
        p[rt++].read();
    }
    sort(p,p+rt);
    m=ConvexHull(p,rt,ch);
    printf("%.0lf
",ANS(ch,m));
}
int main(){
//    fre("in.txt","r",stdin);
    int t=0;
    cin>>t;
    For(i,1,t) printf("Case #%d:
",i),solve();
    return 0;
}