POJ2002 Squares

【题解】

        参考https://blog.csdn.net/ditian1027/article/details/22801635

        用散列表存储点的坐标，并采用开散列方法解决关键码冲突。正方形数量计算方法：对于任意两个点，将这两点作为正方形的一条边的两个端点，构造正方形的另外两个点，随后判断这另外两个点是否在点集中即可。值得注意的是，由于每个正方形有四条边，因此最终计算得到的结果应当除以4。

        另外，开散列方法创造的链表要在每组数据结束时释放，以避免内存泄漏。

【代码】

#include <iostream>
#include <stack>
using namespace std;

#define maxp 1005
#define PRIME 21911
#define OUTOFRANGE 0x7fffffff

struct Point
{
    int x;
    int y;
}pt[maxp];

struct hashTab
{
    int x;
    int y;
    hashTab* next;
    hashTab() {
        x = OUTOFRANGE;
        y = OUTOFRANGE;
        next = NULL;
    }
}ht[PRIME];

void initHashTab()
{
    for (int i = 0; i < PRIME; i++) {
        ht[i].x = OUTOFRANGE;
        ht[i].y = OUTOFRANGE;
        ht[i].next = NULL;
    }
}

void insertHashTab(int x, int y) {
    int key = (x * x + y * y) % PRIME;
    hashTab *htp = &(ht[key]);
    /* Apply open hashing strategy to solve collisions. */
    while (htp->next != NULL) htp = htp->next;
    htp->x = x;
    htp->y = y;
    htp->next = new hashTab();
}

bool pointExist(int x, int y)
{
    int key = (x * x + y * y) % PRIME;
    hashTab *htp = &(ht[key]);
    while (htp->next != NULL) {
        if (htp->x == x && htp->y == y)
            return true;
        htp = htp->next;
    }
    return false;
}

void destroyLinkList()
{
    hashTab *htp = NULL;
    stack<hashTab*> s;
    for (int i = 0; i < PRIME; i++) {
        htp = &(ht[i]);
        while (htp->next != NULL) {
            s.push(htp);
            htp = htp->next;
        }
        while (!s.empty()) {
            htp = s.top();
            delete (htp->next);
            s.pop();
        }
    }
}

int main()
{
    int n, x, y;
    while (cin >> n) {
        if (n == 0)break;
        long cnt = 0;
        initHashTab();
        for (int i = 0; i < n; i++) {
            cin >> x >> y;
            insertHashTab(x, y);
            pt[i].x = x;
            pt[i].y = y;
        }
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++) {
                int dx = pt[j].x - pt[i].x;
                int dy = pt[j].y - pt[i].y;
                int x1 = pt[i].x - dy;
                int y1 = pt[i].y + dx;
                int x2 = x1 + dx;
                int y2 = y1 + dy;
                if (pointExist(x1, y1) && pointExist(x2, y2))
                    cnt++;
                int x3 = pt[i].x + dy;
                int y3 = pt[i].y - dx;
                int x4 = x3 + dx;
                int y4 = y3 + dy;
                if (pointExist(x3, y3) && pointExist(x4, y4))
                    cnt++;
            }
        cnt /= 4;
        printf("%ld
", cnt);
        destroyLinkList();
    }
    return 0;
}