BZOJ1185: [HNOI2007]最小矩形覆盖

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传送门

旋转卡壳。

首先求凸包没什么好商量的。

然后有一个结论，如果存在一个最小的矩形覆盖，那么凸包里必定存在一条边和矩形的边重合。

自己yy一下就好啦，很容易想明白。

然后枚举每条边，移动另外三条边即可。

注意点积，叉积的结合运用什么的。

//BZOJ 1185
//by Cydiater
//2017.1.29
#include <iostream>
#include <map>
#include <ctime>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <iomanip>
#include <algorithm>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
#define ll long long
#define up(i,j,n)	for(int i=j;i<=n;i++)
#define down(i,j,n)	for(int i=j;i>=n;i--)
#define cmax(a,b)	a=max(a,b)
#define cmin(a,b)	a=min(a,b)
#define Vector 		Point
#define db		double
const int MAXN=1e5+5;
const int oo=0x3f3f3f3f;
const db eps=1e-10;
const db PI=3.14159265358979323846;
struct Point{
	db x,y;
	Point(db x=0,db y=0):x(x),y(y){}
};
Vector operator + (Point x,Point y){return Vector(x.x+y.x,x.y+y.y);}
Vector operator - (Point x,Point y){return Vector(x.x-y.x,x.y-y.y);}
Vector operator * (Vector x,db p){return Vector(x.x*p,x.y*p);}
Vector operator / (Vector x,db p){return Vector(x.x/p,x.y/p);}
int dcmp(db x){if(fabs(x)<eps)return 0;else return x<0?-1:1;}
bool operator < (const Vector &x,const Vector &y){return dcmp(x.x-y.x)==0?x.y<y.y:x.x<y.x;}
bool operator == (const Vector &x,const Vector &y){return dcmp(x.x-y.x)==0&&dcmp(x.y-y.y)==0;}
int N,top;
Point V[MAXN],q[MAXN],aim[10];
db ans=oo;
namespace solution{
	void P(Point x){cout<<x.x<<‘ ‘<<x.y<<endl;}
	db Cross(Vector x,Vector y){return x.x*y.y-x.y*y.x;}
	db Dot(Vector x,Vector y){return x.x*y.x+x.y*y.y;}
	db Len(Vector x){return sqrt(Dot(x,x));}
	Point Write(){
		db x,y;scanf("%lf%lf",&x,&y);
		return Point(x,y);
	}
	void Prepare(){
		scanf("%d",&N);
		up(i,1,N)V[i]=Write();
		sort(V+1,V+N+1);
		N=unique(V+1,V+N+1)-(V+1);
	}
	void Andrew(){
		up(i,1,N){
			while(top>=2&&dcmp(Cross(q[top]-q[top-1],V[i]-q[top-1]))<=0)top--;
			q[++top]=V[i];
		}
		int lim=top;
		down(i,N-1,1){
			while(top-lim>=1&&dcmp(Cross(q[top]-q[top-1],V[i]-q[top-1]))<=0)top--;
			q[++top]=V[i];			
		}
	}
	Vector rotate(Vector x,db rad){return Vector(x.x*cos(rad)-x.y*sin(rad),x.x*sin(rad)+x.y*cos(rad));}
	void Solve(){
		Andrew();
		top--;
		int p2=2,p3=3,p4=4;
		up(p1,1,top){
			while(dcmp(Dot(q[p2+1]-q[p2],q[p1]-q[p1+1]))<0){p2%=top;p2++;}
			while(dcmp(Cross(q[p3]-q[p3+1],q[p1+1]-q[p1]))>0){p3%=top;p3++;}
			while(dcmp(Dot(q[p4+1]-q[p4],q[p1+1]-q[p1]))<0){p4%=top;p4++;}
			db bot=fabs(Dot(q[p4]-q[p2],q[p1+1]-q[p1])/Len(q[p1+1]-q[p1]));
			db hig=fabs(Cross(q[p3]-q[p1],q[p1+1]-q[p1]))/Len(q[p1+1]-q[p1]);
			if(bot*hig<ans){
				ans=bot*hig;
				aim[1]=q[p1+1]+(q[p1+1]-q[p1])/Len(q[p1+1]-q[p1])*fabs(Dot(q[p2]-q[p1+1],q[p1+1]-q[p1])/Len(q[p1+1]-q[p1]));
				aim[2]=aim[1]+rotate((q[p1+1]-q[p1])/Len(q[p1+1]-q[p1]),PI/2.0)*hig;
				aim[3]=aim[2]+rotate((q[p1+1]-q[p1])/Len(q[p1+1]-q[p1]),PI)*bot;
				aim[4]=aim[3]+rotate((q[p1+1]-q[p1])/Len(q[p1+1]-q[p1]),PI*1.5)*hig;
			}
		}
		printf("%.5lf\n",ans);
		int pos=0;
		aim[0]=Point(oo,oo);
		up(i,1,4)if(dcmp(aim[i].y-aim[pos].y)<0||(dcmp(aim[i].y-aim[pos].y)==0&&dcmp(aim[i].x-aim[pos].x)<0))pos=i;
		up(i,1,4){
			printf("%.5lf %.5lf\n",dcmp(aim[pos].x)==0?0:aim[pos].x,dcmp(aim[pos].y)==0?0:aim[pos].y);
			pos%=4;pos++;
		}
	}
}
int main(){
	//freopen("input.in","r",stdin);
	using namespace solution;
	Prepare();
	Solve();
	return 0;
}

BZOJ1185: [HNOI2007]最小矩形覆盖

标签：write   eps   n+1   sort   ring   hnoi   php   .com   bsp   

原文地址：http://www.cnblogs.com/Cydiater/p/6357616.html