Poj2002squares (number of squares composed of point sets)

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Two vertices can be enumerated, because the two outer vertices of a square can be obtained by known results. It is said that the following formula can be obtained based on the equi of a triangle. The mathematical scum will not prove it...

Known: (x1, Y1) (X2, Y2)

Then: X3 = X1 + (y1-y2) Y3 = Y1-(x1-x2)

X4 = x2 + (y1-y2) Y4 = Y2-(x1-x2)

Or

X3 = x1-(y1-y2) Y3 = Y1 + (x1-x2)

X4 = x2-(y1-y2) Y4 = y2 + (x1-x2)

Then we can do the hash or binary. Here we only use the hash to do it.

It should be regarded as a simple hash application for resolving conflicts and put it in an adjacent table.

Two vertices need to be enumerated twice to ensure that the squares at both locations are enumerated.

The final result must be divided by 4 because the enumeration is repeated.

 1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set>10 using namespace std;11 #define N 101012 #define mod 9999113 #define LL long long14 #define INF 0xfffffff15 const double eps = 1e-8;16 const double pi = acos(-1.0);17 const double inf = ~0u>>2;18 struct point19 {20     int x,y;21     point(int x=0,int y=0):x(x),y(y){}22 }p[N],o[N];23 int next[N],head[mod],t;24 void insert(int i)25 {26     int key = (p[i].x*p[i].x+p[i].y*p[i].y)%mod;27     next[t] = head[key];28     o[t].x = p[i].x;29     o[t].y = p[i].y;30     head[key] = t++;31 }32 int find(point a)33 {34     int  key = (a.x*a.x+a.y*a.y)%mod;35     int i;36     for(i = head[key] ; i!= -1 ; i = next[i])37     {38         if(o[i].x==a.x&&o[i].y == a.y) return 1;39     }40     return 0;41 }42 bool cmp(point a,point b)43 {44     if(a.x==b.x)45     return a.y<b.y;46     return a.x<b.x;47 }48 int main()49 {50     int n,i,j;51     while(scanf("%d",&n)&&n)52     {53         memset(head,-1,sizeof(head));54         t = 0;55         for(i = 1; i <= n; i++)56         {57             scanf("%d%d",&p[i].x,&p[i].y);58             insert(i);59         }60         //sort(p+1,p+n+1,cmp);61         int ans = 0;62         for(i = 1; i <= n ;i++)63             for(j = 1 ; j <= n; j++)64             {65                 if(i==j) continue;66                 point p1,p2;67                 p1.x = p[i].x+(p[i].y-p[j].y);68                 p1.y = p[i].y-(p[i].x-p[j].x);69                 p2.x = p[j].x+(p[i].y-p[j].y);70                 p2.y = p[j].y-(p[i].x-p[j].x);71                 if(!find(p1)) continue;72                 if(!find(p2)) continue;73                 ans++;74             }75         printf("%d\n",ans/4);76     }77     return 0;

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