POJ3432《Count Squares》方法：哈希

题目大意：跟POJ2002一摸一样，就数据由1000变为2000，找正方形形。

解题思路：已知： (x1,y1)  (x2,y2)

 则：   x3=x1+(y1-y2)   y3= y1-(x1-x2)

   x4=x2+(y1-y2)   y4= y2-(x1-x2)

 或

   x3=x1-(y1-y2)   y3= y1+(x1-x2)

   x4=x2-(y1-y2)   y4= y2+(x1-x2)

// 304k 297ms#include <iostream>#include <cstring>using namespace std;const int prime = 1999;typedef class Node {public:int x, y;} Node;Node node[2000];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;}