BZOJ2458 [BeiJing2011] Minimum triangle

Description

Xaviera is now encountering an interesting question.
There are n points on the plane, Xaviera want to find the smallest triangle in the perimeter.
Because of the very many points, the distribution is also very chaotic, so Xaviera want to ask you to solve this problem.
To reduce the difficulty of the problem, the triangles here also include three points of the collinear line.

Input

The first line contains an integer n representing the number of points.
The next n rows have two integers per line, representing the coordinates of the point.

Output

The output has only one row, containing a 6-bit decimal, which is the perimeter of the triangle with the shortest perimeter (rounded).

Sample Input4
1 1
2 3
3 3
3 4

Sample Output3.414214


HINT

100% of the data are n≤200000.

Source

Day1

Positive solution: Divide and conquer the problem report: with the plane nearest point to the practice is the same, but when looking for a more layer for on it. Do not repeat, the code is as follows:

1 //It's made by jump~2#include <iostream>3#include <cstdlib>4#include <cstring>5#include <cstdio>6#include <cmath>7#include <algorithm>8#include <ctime>9#include <vector>Ten#include <queue> One#include <map> A#include <Set> - using namespacestd; -typedefLong LongLL; the Const intMAXN =200011; - #defineRG Register - Const DoubleINF =1e20; - intN; + Doubleans; - structnode{ +     Doublex, y; A }A[MAXN],B[MAXN]; at  -InlineintGetint () - { -Rgintw=0, q=0;CharC=GetChar (); -      while((c<'0'|| C>'9') && c!='-') C=getchar ();if(c=='-') q=1, c=GetChar (); -      while(c>='0'&& c<='9') w=w*Ten+c-'0', C=getchar ();returnQ? -w:w; in } -InlineBOOLCMP (node Q,node QQ) {returnq.x<qq.x;} toInlineBOOLCmpy (node Q,node QQ) {returnq.y<qq.y;} +InlineDouble Get(RGintI,rgintj) {returnsqrt ((a[i].x-a[j].x) * (a[i].x-a[j].x) + (A[I].Y-A[J].Y) * (a[i].y-a[j].y));} -InlineDoubleGETB (RGintI,rgintj) {returnsqrt ((b[i].x-b[j].x) * (b[i].x-b[j].x) + (B[I].Y-B[J].Y) * (b[i].y-b[j].y));} theInlinevoidDiv_solve (intLintR) { *     if(L==R)return;if(L +1==R)return ; $RgintMid= (L+R)/2; Div_solve (L,mid); Div_solve (mid+1, R);Long DoubleDis1,dis2,dis3;Panax NotoginsengRgintCnt=0; b[++cnt]=A[mid]; -      for(RGinti=mid-1; i>=l;i--)if(A[mid].x-a[i].x<=ans) b[++cnt]=a[i];Else  Break; the      for(RGintI=mid+1; i<=r;i++)if(A[i].x-a[mid].x<=ans) b[++cnt]=a[i];Else  Break; +Sort (b +1, b+cnt+1, cmpy); A      for(RGintI=1; i<cnt;i++) the      for(RGintj=i+1; j<cnt;j++) { +         if(Fabs (B[I].Y-B[J].Y) >ans) Break;  -dis1=Getb (i,j); $          for(RGintk=j+1; k<=cnt;k++) { $         if(Fabs (B[I].Y-B[K].Y) >ans) Break; -DIS2=GETB (J,K); dis3=Getb (i,k); -         if(Dis1+dis2+dis3<ans) ans=dis1+dis2+Dis3; the         } -     }Wuyi } the  -InlinevoidWork () { WuN=getint (); for(RGintI=1; i<=n;i++) A[i].x=getint (), a[i].y=getint (); -Sort (A +1, a+n+1, CMP); ans=Get(n2, N-1)+Get(n2, N) +Get(n1, n); AboutDiv_solve (1, n); $printf"%.6LF", ans); - } -  - intMain () A { + Work (); the     return 0; -}

BZOJ2458 [BeiJing2011] Minimum triangle