POJ 2002 Squares (two minutes)

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 17423		Accepted: 6614

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

Give a flat scatter set to determine how many squares can be formed. Although there are 3.5 seconds, the four-layer cycle of violence will definitely be timed out. So there is a way of thinking: First point sorting, double-loop enumeration before (N-2) points, in order to prevent repeated judgment, the second loop in the J to start from the i+1, two points after the search (N-J) point in the presence of a can and s[i],s[j] form the point of the square, so the second layer loop end condition is j< =n-2, the remaining 2 points are used to find the range of binary lookups is [j+1,n].

Known 2 points, write the coordinates that can form a square with these 2 points, such as (do not use double when calculating coordinates, otherwise it is easy to tle or WA)

#include <cstdio> #include <iostream> #include <algorithm>using namespace std;const int maxn=1e5+20;    int N;long long ans;struct star{int x, y;        Star () {} star (int x,int y) {this->x=x;    this->y=y;        } bool operator< (const star& N) Const {if (this->x==n.x) return this->y<n.y;    Return this->x<n.x;        }}s[maxn];int searchh (int l,int R,star N) {while (l<=r) {int mid= (L+R)/2;        if (S[MID].X==N.X&AMP;&AMP;S[MID].Y==N.Y) return 1;        if (s[mid]<n) l=mid+1;    else r=mid-1; } return 0;} int main () {#ifndef Online_judge freopen ("In.txt", "R", stdin), #endif//Online_judge while (scanf ("%d", &n)!=eof&amp        ; &n) {ans=0;        for (int i=1;i<=n;i++) scanf ("%d%d", &s[i].x,&s[i].y);        Sort (s+1,s+n+1);    for (int i=1;i<=n-3;i++) {for (int j=i+1;j<=n-2;j++) {            int xxx=s[j].x-s[i].x;                int yyy=s[j].y-s[i].y;                int sx1=s[i].x+yyy;                int sy1=s[i].y-xxx;                int sx2=s[j].x+yyy;                int sy2=s[j].y-xxx;                if (Searchh (J+1,n,star (sx1,sy1)) &&searchh (J+1,n,star (sx2,sy2))) ans++;                int sx3=s[i].x-yyy;                int sy3=s[i].y+xxx;                int sx4=s[j].x-yyy;                int sy4=s[j].y+xxx;            if (Searchh (J+1,n,star (sx3,sy3)) &&searchh (J+1,n,star (SX4,SY4))) ans++;    }} printf ("%lld\n", ans); } return 0;}

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POJ 2002 Squares (two minutes)