bzoj1185: [HNOI2007]最小矩形覆盖 计算几何 旋转卡壳

又被卡精度。。。这概率绝了！！！

我们可以肯定凸包的一边肯定在矩形上，所以我们直接寻找上，左，右三个顶点就可以了。

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <cmath>using namespace std;#define eps 1e-10#define maxn 51000struct node{    double x,y;}p[maxn],rec[10];double ans=1000000000000;int n;double sqr(double x){    return x*x;}double dis(node a,node b){    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}node operator -(node a,node b){    node t;    t.x=a.x-b.x;t.y=a.y-b.y;    return t;}node operator +(node a,node b){    node t;    t.x=a.x+b.x;t.y=a.y+b.y;    return t;}double operator *(node a,node b){    return a.x*b.y-a.y*b.x;}bool operator <(node a,node b){    if(fabs(a.y-b.y)<eps) return a.x<b.x;    return a.y<b.y;}bool cmp(node a,node b){    double t=(a-p[1])*(b-p[1]);    if(fabs(t)<=eps) return dis(a,p[1])<dis(b,p[1]);    return t>0;}node operator*(node a,double b){    node tmp;tmp.x=a.x*b;tmp.y=a.y*b;    return tmp;}double operator/(node a,node b){    return a.x*b.x+a.y*b.y;}node s[maxn];int top;void graham(){    for(int i=2;i<=n;i++) if(p[i]<p[1])swap(p[i],p[1]);    sort(p+2,p+n+1,cmp);    s[1]=p[1];s[2]=p[2];top=2;    for(int i=3;i<=n;i++)    {        for(;top>1 && (s[top]-s[top-1])*(p[i]-s[top-1])<=0;top--);        top++;s[top]=p[i];    }    s[top+1]=p[1];    //printf("%d\n",top);}double get(int a,int b,int c){    double tmp=(s[a]-s[b])*(s[a]-s[c]);    return fabs(tmp);}double check(int a,int b,int c){    return (s[a]-s[b])/(s[c]-s[b]);}void rc(){    int l=1,r=1,m=1;    double L,R,D,H;    for(int i=1;i<=top;i++)    {        D=dis(s[i],s[i+1]);        while(get(m+1,i,i+1)-get(m,i,i+1)>-eps ) m=m%top+1;        while(check(r+1,i,i+1)-check(r,i,i+1)>-eps) r=r%top+1;        if(i==1) l=r;        while(check(l+1,i+1,i)-check(l,i+1,i)>-eps) l=l%top+1;        L=(s[i+1]-s[i])/(s[l]-s[i])/D;        R=(s[i+1]-s[i])/(s[r]-s[i])/D;H=fabs( (s[i+1]-s[i])*(s[m]-s[i])/D );double tmp=(R-L)*H;if(tmp<ans){    ans=tmp;rec[0]=s[i]+(s[i+1]-s[i])*(R/D);rec[1]=rec[0]+(s[r]-rec[0])*(H/dis(rec[0],s[r]));rec[2]=rec[1]-(rec[0]-s[i])*((R-L)/dis(s[i],rec[0]));rec[3]=rec[2]-(rec[1]-rec[0]);}    }}int main(){    scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);graham();rc();printf("%.5lf\n",ans);int tmp=0;for(int i=1;i<=3;i++)if(rec[i]<rec[tmp])tmp=i;for(int i=0;i<=3;i++)printf("%.5lf %.5lf\n",rec[(i+tmp)%4].x,rec[(i+tmp)%4].y);    return 0;}/*61 31 42 13 03 66 3*/