BZOJ1337: Minimum Circle cover

Title: Ask for the minimum circle coverage of n points.

Puzzle: Minimum circle overlay, upper template. Proof of complexity can be stamped: here

Code:

1#include <cstdio>2#include <cstdlib>3#include <cmath>4#include <cstring>5#include <algorithm>6#include <iostream>7#include <vector>8#include <map>9#include <Set>Ten#include <queue> One#include <string> A #defineINF 1000000000 - #defineMAXN 1000000+5 - #defineMAXM 200000+5 the #defineEPS 1e-6 - #definell Long Long - #defineull unsigned long Long - #definePA pair<int,int> + #defineFor0 (i,n) for (int i=0;i<= (n); i++) - #defineFor1 (i,n) for (int i=1;i<= (n); i++) + #defineFor2 (i,x,y) for (int i= (x); i<= (y); i++) A #defineFor3 (i,x,y) for (int i= (x); i>= (y); i--) at #defineFor4 (i,x) for (int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go) - #defineFor5 (N,M) for (int. i=1;i<=n;i++) for (int j=1;j<=m;j++) - #defineMoD 1000000007 - #defineLCH K<<1,l,mid - #defineRCH K<<1|1,mid+1,r - #defineSQR (x) (x) * (x) in #defineDB Double - using namespacestd; toInlineintRead () + { -     intx=0, f=1;CharCh=GetChar (); the      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} *      while(ch>='0'&&ch<='9') {x=Ten*x+ch-'0'; ch=GetChar ();} $     returnx*F;Panax Notoginseng } - struct Point the { + db x, y; APointoperator+ (point B) {return(point) {x+b.x,y+b.y};} thePointoperator/(db b) {return(point) {x/b,y/b};} + }A[MAXN]; - intN; $ db ans; $Inline DB Dist (point A,point b) {returnsqrt (SQR (a.x-b.x) +SQR (a.y-b.y));} - Inline Point Calc (db a,db b,db c,db d,db e,db f) - { the   return(point) {(c*e-f*b)/(A*e-d*b), (a*f-c*d)/(a*e-d*b)}; - }WuyiInline pointGet(Point a,point b,point c) the { -    returnCalc (b.x-a.x,b.y-a.y, (Sqr (b.x) +sqr (B.Y)-sqr (a.x)-SQR (A.Y))/2.0, WuC.x-b.x,c.y-b.y, (Sqr (c.x) +sqr (C.Y)-sqr (b.x)-SQR (B.Y))/2.0); - } About intMain () $ { -Freopen ("Input.txt","R", stdin); -Freopen ("output.txt","W", stdout); -n=read (); AFor1 (i,n) scanf ("%LF%LF",&a[i].x,&a[i].y); +For1 (i,n) Swap (A[rand ()%n+1],a[rand ()%n+1]); theans=0; -For1 (I,n)if(Dist (a[i],a[0]) >ans+EPS) $    { thea[0]=a[i];ans=0; theFor1 (j,i-1)if(Dist (a[j],a[0]) >ans+EPS) the         { thea[0]= (A[i]+a[j])/2; Ans=dist (a[0],a[i]); -For1 (k,j-1)if(Dist (a[k],a[0]) >ans+EPS) in             { thea[0]=Get(A[i],a[j],a[k]); Ans=dist (a[0],a[i]); the             } About         } the    } theprintf"%.2f%.2f%.2f\n", a[0].x,a[0].y,ans); the    return 0; +}

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I'm not going to tell you this is three times times the experience.

BZOJ1337: Minimum Circle cover