Bzoj1185 [HNOI2007]最小矩形覆盖

Time Limit: 10 Sec  Memory Limit: 162 MBSec  Special Judge
Submit: 1653  Solved: 745
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计算几何 vfleaking提供Spj

 

数学问题 计算几何 旋转卡壳

按顺序枚举凸包上每一条边为底边，旋转卡壳找出此时的矩形的顶点、最左点和最右点，花式计算出边长，求面积，记录面积最小值和对应的解。

式子都能看懂，但不看题解就是写不出来，沮丧。

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const int mxn=100010;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline int DT(double x){
    if(abs(x)<eps)return 0;
    else return x<0?-1:1;
}
struct point{
    double x,y;
    point operator + (point b){return (point){x+b.x,y+b.y};}
    point operator - (point b){return (point){x-b.x,y-b.y};}
    point operator * (double v){return (point){x*v,y*v};}
    point operator / (double v){return (point){x/v,y/v};}
    bool operator < (point b)const{
        return x<b.x || (x==b.x && y<b.y);
    }
}a[mxn],p[mxn],ans[5];
typedef point Vector;
double dot (point a,point b){return a.x*b.x+a.y*b.y;}
double Cross(point a,point b){
    return a.x*b.y-a.y*b.x;
}
double dist(point a,point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double Len(point a){
    return sqrt((a.x*a.x)+(a.y*a.y));
}
double DLC(point p,point a,point b){
    Vector v1=p-a,v2=b-a;
    return fabs(Cross(v1,v2)/Len(v2));
}
int n;
double ANS=0;
int st[mxn],top=0,t2=0;
void Andrew(){
    sort(p+1,p+n+1);
    for(int i=1;i<=n;i++){
        while(top>1 && Cross(p[st[top]]-p[st[top-1]],p[i]-p[st[top-1]])<=0)top--;
        st[++top]=i;
    }
    t2=top;
    for(int i=n-1;i;i--){
        while(t2>top && Cross(p[st[t2]]-p[st[t2-1]],p[i]-p[st[t2-1]])<=0)t2--;
        st[++t2]=i;
    }
    if(n>1)t2--;
    n=t2;
    for(int i=1;i<=n;i++){
        a[i]=p[st[i]];
    }
    return;
}
void solve(){
    ANS=(double)INF;
    a[n+1]=a[1];
    int l=1,r=1,tmp=1;
    double H,D,L,R;
    for(int i=1;i<=n;i++){
        while(DT(DLC(a[tmp],a[i],a[i+1])-DLC(a[tmp+1],a[i],a[i+1]))<=0)tmp=tmp%n+1;//顶部 
        while(DT(dot(a[i+1]-a[i],a[l]-a[i])-dot(a[i+1]-a[i],a[l+1]-a[i]))<=0)l=l%n+1;//左边 
        if(i==1)r=tmp;
        while(DT(dot(a[i+1]-a[i],a[r]-a[i])-dot(a[i+1]-a[i],a[r+1]-a[i]))>=0)r=r%n+1;//右边 
//      printf("i:%d l:%d tmp:%d r:%d\n",i,l,tmp,r);
        L=dist(a[i+1],a[i]);
//      H=fabs(Cross(a[tmp]-a[i],a[i+1]-a[i]))/L;//高 
        H=DLC(a[tmp],a[i],a[i+1]);
        D=fabs(dot(a[i+1]-a[i],a[l]-a[i])/L)+fabs(dot(a[i+1]-a[i],a[r]-a[i])/L);//底边长 
        R=H*D;
//      printf("L:%.3f H:%.3f D:%.3f R:%.3f\n",L,H,D,R);
        if(R<ANS){//更新答案 
            ANS=R;
            Vector v=(a[i+1]-a[i])/L;//单位长度
            Vector u=(point){-v.y,v.x};//v逆时针转90度 
            ans[1]=a[i]+v*dot(a[r]-a[i],v);
            ans[2]=a[i]+v*dot(a[l]-a[i],v);
            ans[3]=ans[2]+u*H;
            ans[4]=ans[1]+u*H;
        }
    }
    return;
}
int main(){
//  freopen("in.txt","r",stdin);
    int i,j;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%lf%lf",&p[i].x,&p[i].y);
    }
    Andrew();
/*  printf("n:%d\n",n);
    for(i=1;i<=n;i++){
        printf("%.3f  %.3f\n",a[i].x,a[i].y);
    }*/
    solve();
    int pos=1;
    for(i=1;i<=4;i++){
        if(ans[i].y<eps)ans[i].y=0;
        if(ans[i].x<eps)ans[i].x=0;
        if(ans[i].y<ans[pos].y || (ans[i].y==ans[pos].y && ans[i].x<ans[pos].x))pos=i;
    }
    if(ANS<eps)ANS=0;
    printf("%.5f\n",ANS);
    for(i=pos;i<=4;i++)
        printf("%.5f %.5f\n",ans[i].x,ans[i].y);
    for(i=1;i<pos;i++)
        printf("%.5f %.5f\n",ans[i].x,ans[i].y);
    return 0;
}