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。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<queue>
  6 #include<set>
  7 #include<map>
  8 #include<stack>
  9 #include<time.h>
 10 #include<cstdlib>
 11 #include<cmath>
 12 #include<list>
 13 using namespace std;
 14 #define MAXN 100100
 15 #define eps 1e-9
 16 #define For(i,a,b) for(int i=a;i<=b;i++) 
 17 #define Fore(i,a,b) for(int i=a;i>=b;i--) 
 18 #define lson l,mid,rt<<1
 19 #define rson mid+1,r,rt<<1|1
 20 #define mkp make_pair
 21 #define pb push_back
 22 #define cr clear()
 23 #define sz size()
 24 #define met(a,b) memset(a,b,sizeof(a))
 25 #define iossy ios::sync_with_stdio(false)
 26 #define fre freopen
 27 #define pi acos(-1.0)
 28 #define inf 1e6+7
 29 #define Vector Point
 30 const int Mod=1e9+7;
 31 typedef unsigned long long ull;
 32 typedef long long ll;
 33 int dcmp(double x){
 34     if(fabs(x)<=eps) return 0;
 35     return x<0?-1:1;
 36 }
 37 struct Point{
 38     double x,y;
 39     Point(double x=0,double y=0):x(x),y(y) {}
 40     bool operator < (const Point &a)const{
 41         if(x==a.x) return y<a.y;
 42         return x<a.x;
 43     }
 44     Point operator - (const Point &a)const{
 45         return Point(x-a.x,y-a.y);
 46     }
 47     Point operator + (const Point &a)const{
 48         return Point(x+a.x,y+a.y);
 49     }
 50     Point operator * (const double &a)const{
 51         return Point(x*a,y*a);
 52     }
 53     Point operator / (const double &a)const{
 54         return Point(x/a,y/a);
 55     }
 56     void read(){
 57         scanf("%lf%lf",&x,&y);
 58     }
 59     void out(){
 60         cout<<"debug: "<<x<<" "<<y<<endl;
 61     }
 62     bool operator == (const Point &a)const{
 63         return dcmp(x-a.x)==0 && dcmp(y-a.y)==0;
 64     }
 65 };
 66 double Dot(Vector a,Vector b) {
 67     return a.x*b.x+a.y*b.y;
 68 }
 69 double dis(Vector a) {
 70     return sqrt(Dot(a,a));
 71 }
 72 double Cross(Point a,Point b){
 73     return a.x*b.y-a.y*b.x;
 74 }
 75 int ConvexHull(Point *p,int n,Point *ch){
 76     int m=0;
 77     For(i,0,n-1) {
 78         while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
 79         ch[m++]=p[i];
 80     }
 81     int k=m;
 82     Fore(i,n-2,0){
 83         while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
 84         ch[m++]=p[i];
 85     }
 86     if(n>1) m--;
 87     return m;
 88 }
 89 double ANS(Point *p,int n){
 90     int L,R=1,U=1;
 91     double ans=1e9+9;
 92     p[n]=p[0];
 93     For(i,0,n-1) {
 94         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;
 95         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;
 96         if(!i) L=R;
 97         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;
 98         double tmp=fabs(Cross(p[U]-p[i],p[i+1]-p[i]))/dis(p[i+1]-p[i]);
 99         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]);
100         ans=min(ans,(cnt2+cnt1)*tmp);
101     }
102     return ans;
103 }
104 int n,m;
105 Point p[200005];
106 Point ch[200005];
107 void solve(){
108     scanf("%d",&n);
109     int rt=0;
110     For(i,0,n-1) {
111         p[rt++].read();
112         p[rt++].read();
113         p[rt++].read();
114         p[rt++].read();
115     }
116     sort(p,p+rt);
117     m=ConvexHull(p,rt,ch);
118     printf("%.0lf\n",ANS(ch,m));
119 }
120 int main(){
121 //    fre("in.txt","r",stdin);
122     int t=0;
123     cin>>t;
124     For(i,1,t) printf("Case #%d:\n",i),solve();
125     return 0;
126 }
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