poj2002 (very simple math/geometry +hash)

Links: Http://poj.org/problem?id=2002Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 21720		Accepted: 8321

Description

A square is a 4-sided polygon whose sides has equal length and adjacent sides form 90-degree angles. It is also a polygon such the degrees gives the same polygon of It centre by. It isn't the only polygon with the latter property, however, as a regular octagon also have this property.

So we are know what's a square looks like, but can we find all possible squares that can is formed from a set of stars in a Night sky? To make the problem easier, we'll assume that the night sky was a 2-dimensional plane, and each star was specified by its X and Y coordinates.

Input

The input consists of a number of test cases. Each test case is starts with the integer n (1 <= n <=) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (both integers) of each point. Assume that the points is distinct and the magnitudes of the coordinates is less than 20000. The input is terminated when n = 0.

Output

For each test case, print on a line the number of squares one can form from the given stars.

Sample Input

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

Let's leave a few pits:

① is said to have a two-part approach, which accounts for more pit-waiting;

② started with a vector, t ... Turn the vector into an array, then optimize the constant, occupy the pit for more

Main topic:

There are N (1<=n<=1000) points in the plane, coordinates of the given point xi,yi, how many squares can be formed altogether

Analysis: According to the relationship between the relations, from which two points a, a and two points to determine the m,n, see if m,n exists, determine the time complexity of two points is O (N2), the time complexity of the lookup is O (logn), the overall O (2N2LOGN) data range within 1000, acceptable

Start with the vector match, determine the N2 bar vector, and then determine the starting point and size of a vector, direction (parallel), vector matching, and then T (Pit ①) (there is a (wrong) idea, two vectors perpendicular, but this can only determine three points, no longer think) and then based on two points to match the other two points , 900+ms,a dropped (it seems that the constant is not written)

On the code:

1#include <iostream>2#include <cstdio>3#include <queue>4#include <cstdio>5#include <cstring>6#include <cstdlib>7     #defineMem (A, B) memset (A,b,sizeof (a))8     using namespacestd;9 Ten     Const intMod=100007; One     Const intmaxn=120009; A     structnode -     { -         intX,y,next; the }EDGE[MAXN]; -     intCNT; -     intANS,HEAD[MAXN]; -     intxi[1111],yi[1111]; +InlineintRead () -     { +         Charch = getchar ();intx =0, F =1; A          while(Ch <'0'|| CH >'9') {if(ch = ='-') F =-1; CH =GetChar ();} at          while('0'<= CH && Ch <='9') {x = x *Ten+ CH-'0'; CH =GetChar ();} -         returnX *F; -     } -     void  out(intA//instead of printf, faster, commonly known as output hanging -     { -         if(A >9) in         { -              out(ATen); to         } +Putchar (a%Ten+'0'); -     } the     voidInsintXintYinth) *     { $h%=MoD;Panax Notoginsengedge[cnt].x=x; -edge[cnt].y=y; theedge[cnt].next=Head[h]; +head[h]=cnt++; A     } theInlineBOOLCMP (node A,node b) +     { -         if(A.X==B.X&AMP;&AMP;A.Y==B.Y)return 1; $         return 0; $     } -     BOOLSearch (Node A,inth) -     { theh%=MoD; -         //printf ("search:\n");Wuyi          for(inti=head[h];i!=-1; i=edge[i].next) the         { -             //"%d%d%d%d%d%d%d%d%d%d%d %d\n", xi[i],yi[i],xi[j],yi[j],a.x,a.y,b.x,b.y,c.x,c.y,d.x,d.y);  Wu             if(CMP (EDGE[I],A))return 1; -         } About         return 0; $     } -     voidInit () -     { -Mem (Edge,0); AMEM (head,-1); +Mem (xi,0); theMem (Yi,0); -Cnt=0; $ans=0; the     } the     intMain () the     { the         intN; -          while(SCANF ("%d", &n) &&N) in         { the init (); the              for(intI=1; i<=n;i++) About             { theXi[i]=read (); yi[i]=read (); theIns (Xi[i],yi[i],abs (xi[i]* ++yi[i])); the             } +              for(intI=1; i<=n;i++) -             { the                  for(intj=i+1; j<=n;j++)Bayi                 { the                     intdty=yi[j]-yi[i],dtx=xi[j]-Xi[i]; the node a,b,c,d; -a.x=xi[i]+dty;a.y=yi[i]-DTX; -b.x=xi[j]+dty;b.y=yi[j]-DTX; thec.x=xi[i]-dty;c.y=yi[i]+DTX; thed.x=xi[j]-dty;d.y=yi[j]+DTX; the                     intHa=abs (a.x* ++A.Y), Hb=abs (b.x* ++b.y), theHc=abs (c.x* ++C.Y), Hd=abs (d.x* ++d.y); -                     //printf ("%d%d%d%d%d%d%d%d\n%d%d%d ", xi[i],yi[i],xi[j],yi[j],a.x,a.y,b.x,b.y,c.x,c.y,d.x,d . y); the                     if(Search (A,ha) &&search (B,HB)) {ans++;} the                     if(Search (C,HC) &&search (D,HD)) {ans++;} the                 }94             } the              out(ans/4);p Utchar ('\ n'); the         } the}

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poj2002 (very simple math/geometry +hash)