矩形面积

Accepts: 717

Submissions: 1619

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Problem Description

小度熊有一个桌面，小度熊剪了很多矩形放在桌面上，小度熊想知道能把这些矩形包围起来的面积最小的矩形的面积是多少。

Input

第一行一个正整数 T，代表测试数据组数(1≤T≤20 )，接下来
T 组测试数据。

每组测试数据占若干行，第一行一个正整数 N(1≤N<≤1000) ，代表矩形的数量。接下来
N 行，每行 8 个整数x 1 ,y 1 ,x 2 ,y 2 ,x 3 ,y 3 ,x 4 ,y 4  ，代表矩形的四个点坐标，坐标绝对值不会超过10000。

Output

对于每组测试数据，输出两行：

第一行输出"Case #i:"，i 代表第 i 组测试数据。 第二行包含1 个数字，代表面积最小的矩形的面积，结果保留到整数位。

Sample Input

2
2
5 10 5 8 3 10 3 8
8 8 8 6 7 8 7 6
1
0 0 2 2 2 0 0 2

Sample Output

Case #1:
17
Case #2:
4

旋转卡壳，和bzoj上那题基本相同

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define MAXN (10000+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;   }
};
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()
{
//	freopen("F.in","r",stdin);

	int T;
	cin>>T;
	For(kcase,T)
	{
		ans=INF;
	    scanf("%d",&n);
	    n*=4;
	    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("Case #%d:\n",kcase);
	    printf("%.0lf\n",ans);    

	}
    return 0;
}