POJ2002《Squares》方法：哈希

题目大意：给定平面上的一些座标点(1000个)，找出这些点能构成多少个正方形，座标取值不大于20000。

解题思路：若遍历4个点的座标，肯定超时。因此先取两个点座标，然后用哈希值查找正方形的另两个点座标，key = (x*x+y*y)%prime，这里prime取1999(不大于2n的素数)，并用链地址法解决地址冲突。因为每个正方形按不同顺序被枚举了4次，所以要除以4。

// 620k 1360ms #include <iostream> #include <cstring> using namespace std; const int prime = 1999; typedef class Node { public: int x, y; } Node; Node node[1000]; typedef class HashTable { public: int x, y; HashTable *next; HashTable() { next = NULL; } } HashTable; HashTable *hashs[prime]; void insert_hash(int x, int y) { int key = (x*x + y*y) % prime; if (!hashs[key]) { HashTable *temp = new HashTable; temp->x = x; temp->y = y; hashs[key] = temp; } else { HashTable *temp = hashs[key]; while (temp->next) { temp = temp->next; } temp->next = new HashTable; temp->next->x = x; temp->next->y = y; } } bool find(int x, int y) { int key = (x*x + y*y) %prime; if (!hashs[key]) { return false; } else { HashTable *temp = hashs[key]; while (temp) { if (temp->x == x && temp->y == y) { return true; } temp = temp->next; } } return false; } int main() { //freopen("temp.txt", "r", stdin); int n, num; while (cin >> n && n != 0) { memset(hashs, 0, sizeof(hashs)); num = 0; for (int i = 0; i < n; ++i) { cin >> node[i].x >> node[i].y; insert_hash(node[i].x, node[i].y); } int x1, x2, x3, x4, y1, y2, y3, y4, a, b; for (int i = 0; i < n-1; ++i) { for (int j = i+1; j < n; ++j) { x1 = node[i].x; y1 = node[i].y; x2 = node[j].x; y2 = node[j].y; a = x1 - x2; b = y1 - y2; x3 = x1 + b; y3 = y1 - a; x4 = x2 + b; y4 = y2 - a; if (find(x3, y3) && find(x4, y4)) { ++num; } x3 = x1 - b; y3 = y1 + a; x4 = x2 - b; y4 = y2 + a; if (find(x3, y3) && find(x4, y4)) { ++num; } } } cout << num/4 << endl; } return 0; }