[BZOJ1185]最小矩形覆盖

上初三后遇到许多杂事，一度想放弃OI。

然后我就想着省赛随便浪，省赛之前沉迷于几乎不考的计算几何(因为写起来爽啊)

这个是写完模板后的第一题，看了之后感觉思路还挺清晰的。

首先因为它的tag是‘凸包‘，所以我们当然要先求凸包啦~Graham就好了

然后有一个结论是矩形的某一条边一定与凸包的某一条边共线，脑补一下如果不共线，稍微转一下矩形就会更小了。

那么我们枚举每一条凸包上的边x，每次找离x最远的点。

这里可以用二分，只有最远的那个点连接的两条边相对于x一个左转一个右转，把这条性质作为二分依据，就可以愉快地二分啦。

这题的细节处理较为烦人，像我这种弱智调了几天才调好。

代码如下

  1 #include<stdio.h>
  2 #include<math.h>
  3 #include<algorithm>
  4 using namespace std;
  5 #define dmax 1.7976931348623158e+308
  6 #define eps 1e-10
  7 struct point{
  8     double x,y,ang;
  9     int num;
 10 }a[100010],t;
 11 struct point2{
 12     double x,y;
 13     point2(){x=y=0;}
 14 }ha[100010];
 15 int n,i,ori,stack[100010];
 16 double miny,minx,orix,oriy,area;
 17 point2 p1,p2,p3,p4;
 18 double bigger(double x,double y){return x-y>=eps;}
 19 double notsm(double x,double y){return bigger(x,y)||abs(x-y)<=eps;}
 20 double angle(double x,double y){
 21     if(abs(x)<=eps)return dmax;
 22     return y/x;
 23 }
 24 bool cmp(point a,point b){
 25     if(bigger(b.ang,a.ang))return 1;
 26     if(bigger(a.ang,b.ang))return 0;
 27     return bigger(b.x*b.x+b.y*b.y,a.x*a.x+a.y*b.y);
 28 }
 29 double cross(double ax,double ay,double bx,double by){
 30     return ax*by-ay*bx;
 31 }
 32 int hassium(double vx,double vy,int ll,int rr){
 33     int l,r,m;
 34     l=ll+1;
 35     r=rr-1;
 36     while(l<=r){
 37         m=(l+r)/2;
 38         orix=cross(vx,vy,ha[m].x-ha[m-1].x,ha[m].y-ha[m-1].y);
 39         oriy=cross(vx,vy,ha[m+1].x-ha[m].x,ha[m+1].y-ha[m].y);
 40         if(orix>=eps){
 41             if(oriy>=eps)l=m+1;
 42             if(notsm(-oriy,0))return m;
 43         }
 44         if(abs(orix)<=eps)return m;
 45         if(orix<-eps)r=m-1;
 46     }
 47 }
 48 point2 getpoint(double xa,double ya,double x1,double y1,double xb,double yb,double x2,double y2){
 49     point2 p;
 50     if(abs(x1)<=eps){
 51         p.x=xa;
 52         p.y=yb;
 53         return p;
 54     }
 55     if(abs(y1)<=eps){
 56         p.x=xb;
 57         p.y=ya;
 58         return p;
 59     }
 60     p.y=((xb-xa)/x1/x2-yb/x1/y2+ya/x2/y1)/(1./x2/y1-1./x1/y2);
 61     p.x=(p.y-ya)/y1*x1+xa;
 62     return p;
 63 }
 64 double recta(double ax,double ay,double bx,double by){
 65     return sqrt((ax*ax+ay*ay)*(bx*bx+by*by));
 66 }
 67 void getrect(double vx,double vy,int num){
 68     int farest=hassium(vx,vy,num,num+stack[0]-1);
 69     int leftest=hassium(-vy,vx,farest,num+stack[0]-1),rightest=hassium(vy,-vx,num,farest);
 70     point2 t1=getpoint(ha[num].x-vx,ha[num].y-vy,vx,vy,ha[rightest].x,ha[rightest].y,vy,-vx),
 71            t2=getpoint(ha[rightest].x,ha[rightest].y,-vy,vx,ha[farest].x,ha[farest].y,vx,vy),
 72            t3=getpoint(ha[leftest].x,ha[leftest].y,-vy,vx,ha[farest].x,ha[farest].y,-vx,-vy),
 73            t4=getpoint(ha[leftest].x,ha[leftest].y,vy,-vx,ha[num].x,ha[num].y,-vx,-vy);
 74     double tarea=recta(t2.x-t1.x,t2.y-t1.y,t4.x-t1.x,t4.y-t1.y);
 75     if(bigger(area,tarea)){
 76         area=tarea;
 77         p1=t1;
 78         p2=t2;
 79         p3=t3;
 80         p4=t4;
 81     }
 82 }
 83 bool comp(point2 a,point2 b,point2 c,point2 d){
 84     return((bigger(b.y,a.y))||(abs(a.y-b.y)<=eps&&bigger(b.x,a.x)))
 85         &&((bigger(c.y,a.y))||(abs(a.y-c.y)<=eps&&bigger(c.x,a.x)))
 86         &&((bigger(d.y,a.y))||(abs(a.y-d.y)<=eps&&bigger(d.x,a.x)));
 87 }
 88 void gao(point2&a){
 89     if(a.x<0&&abs(a.x)<=1e-5)a.x=-a.x;
 90     if(a.y<0&&abs(a.y)<=1e-5)a.y=-a.y;
 91 }
 92 int main(){
 93     scanf("%d",&n);
 94     miny=minx=dmax;
 95     for(i=1;i<=n;i++){
 96         scanf("%lf%lf",&a[i].x,&a[i].y);
 97         a[i].num=i;
 98         if((a[i].x==minx&&a[i].y<miny)||a[i].x<minx){
 99             miny=a[i].y;
100             minx=a[i].x;
101             ori=i;
102         }
103     }
104     orix=a[ori].x;
105     oriy=a[ori].y;
106     for(i=1;i<=n;i++){
107         a[i].x-=orix;
108         a[i].y-=oriy;
109         if(i!=ori)
110          a[i].ang=angle(a[i].x,a[i].y);
111     }
112     t=a[1];
113     a[1]=a[ori];
114     a[ori]=t;
115     sort(a+2,a+n+1,cmp);
116     stack[0]=2;
117     stack[1]=1;
118     stack[2]=2;
119     for(i=3;i<=n;i++){
120         while(stack[0]>1&&cross(a[stack[stack[0]]].x-a[stack[stack[0]-1]].x,a[stack[stack[0]]].y-a[stack[stack[0]-1]].y,
121                                 a[i].x-a[stack[stack[0]]].x,a[i].y-a[stack[stack[0]]].y)<=0)stack[0]--;
122         stack[0]++;
123         stack[stack[0]]=i;
124     }
125     for(i=1;i<=n;i++){
126         a[i].x+=orix;
127         a[i].y+=oriy;
128     }
129     for(i=1;i<=stack[0];i++){
130         ha[i].x=a[stack[i]].x;
131         ha[i].y=a[stack[i]].y;
132     }
133     area=dmax;
134     getrect(ha[1].x-ha[stack[0]].x,ha[1].y-ha[stack[0]].y,1);
135     for(i=1;i<stack[0];i++){
136         ha[i+stack[0]]=ha[i];
137         getrect(ha[i+1].x-ha[i].x,ha[i+1].y-ha[i].y,i+1);
138     }
139     gao(p1);
140     gao(p2);
141     gao(p3);
142     gao(p4);
143     printf("%.5lf\n",area);
144     if(comp(p1,p2,p3,p4))
145      printf("%.5lf %.5lf\n%.5lf %.5lf\n%.5lf %.5lf\n%.5lf %.5lf\n",p1.x,p1.y,p2.x,p2.y,p3.x,p3.y,p4.x,p4.y);
146     if(comp(p2,p1,p3,p4))
147      printf("%.5lf %.5lf\n%.5lf %.5lf\n%.5lf %.5lf\n%.5lf %.5lf\n",p2.x,p2.y,p3.x,p3.y,p4.x,p4.y,p1.x,p1.y);
148     if(comp(p3,p2,p1,p4))
149      printf("%.5lf %.5lf\n%.5lf %.5lf\n%.5lf %.5lf\n%.5lf %.5lf\n",p3.x,p3.y,p4.x,p4.y,p1.x,p1.y,p2.x,p2.y);
150     if(comp(p4,p2,p3,p1))
151      printf("%.5lf %.5lf\n%.5lf %.5lf\n%.5lf %.5lf\n%.5lf %.5lf\n",p4.x,p4.y,p1.x,p1.y,p2.x,p2.y,p3.x,p3.y);
152 }