Count Squares HDU 3432

题目链接:

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Given a set of points with integer coordinates xi, yi, i = 1...N, your program must find all the squares having each of four vertices in one of these points.

Input

Input file contains integer N followed by N pairs of integers xi yi.Constraints-104 ≤ xi, yi ≤ 104, 1 ≤ N ≤ 2000. All points in the input are different.

Output

Output file must contain a single integer — number of squares found.

Sample Input

Sample input 14 0 0 4 3 -3 4 1 7Sample input 291 1  1 2  1 3  2 1  2 2  2 3  3 1  3 2  3 3

Sample Output

Sample output 11Sample output 26

Hint

Bold texts appearing in the sample sections are informative and do not form part of the actual data. 

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题意:

给你 n 个点,判断存在多少个正方形.

思路:

枚举两个点 || 正方形的边, 然后根据枚举的两个点算出未知的两个点,然后在给出的点中查找这两个点是否出现,如果出现则说明存在正方形,反之说明不存在.

查找的话可以使用hash 或者 map, 但是map 我没有试过,不知道具体能不能行.

代码:

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>using namespace std;const int prime = 1999;typedef class{public:    int x,y;}node;typedef class Hash{//手动写hash表public:    int x, y;    Hash *next;    Hash(){        next = 0;    }}Hash;node a[2000+10];Hash *has[prime];void insert_hash(int pos){//hash的对应法则,我们采用平方和    int k = (a[pos].x * a[pos].x + a[pos].y * a[pos].y) % prime + 1;    if(!has[k]){        Hash *tmp = new Hash;        tmp->x = a[pos].x;        tmp->y = a[pos].y;        has[k] = tmp;    }else {        Hash * tmp = has[k];//如果这个点出现过        while(tmp->next)            tmp = tmp->next;        tmp->next = new Hash;        tmp->next->x = a[pos].x;        tmp->next->y = a[pos].y;    }}bool Find(int x, int y){    int  k = (x * x + y * y) % prime + 1;    if(!has[k]) return false;    Hash * tmp = has[k];    while(tmp){        if(tmp->x == x && tmp->y == y)            return true;        tmp = tmp->next;    }return false;}int main(){    int n;    while(scanf("%d", &n) != EOF && n){        memset(has,0, sizeof(has));        for(int  i = 1; i <= n; ++i){            scanf("%d %d", &a[i].x, &a[i].y);            insert_hash(i);        }int num = 0;        for(int i = 1; i <= n; ++i)            for(int j = i+1; j <= n; ++j){                int x = a[j].x - a[i].x;//根据已知点算正方形未知点的坐标                int y = a[j].y - a[i].y;                int x1 = y + a[i].x;                int y1 = a[i].y - x;                int x2 = y + a[j].x;                int y2 = a[j].y - x;                if(Find(x1,y1) && Find(x2,y2))                    ++num;                x1 = a[i].x - y;                y1 = a[i].y + x;                x2 = a[j].x - y;                y2 = a[j].y + x;                if (Find(x1,y1) && Find(x2,y2))                    ++num;            }printf("%d\n",num/4);//由于每个正方形被算了四次,所以除以4    }return 0;}