Squares

---

Squares原题：

---

Description

      A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.

      So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.

Input

      The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.

Output

      For each test case, print on a line the number of squares one can form from the given stars.

我的理解：

---

      本题主要在搜索，找点，所以采用了hash表，并用链地址法解决冲突。

      注意，排序及条件判断解决重复计数。

 

AC代码：

// Squares: hash#include <iostream>#include <vector>#include <list>const int MAX_HASH = 10011;const int MAX_COUNT = 1001;struct point{int x;int y;};std::vector<point> resource(MAX_COUNT);int hashTable[MAX_HASH];std::vector<int> hashChain(MAX_COUNT);int Hash(int x){int h = x % MAX_HASH;if (h < 0)h += MAX_HASH;return h;}void Insert(point p){int n = Hash(p.x * p.x + p.y * p.y);bool flag = false;for (int i = hashTable[n]; i != -1; i = hashChain[i]){if (resource[i].x == p.x && resource[i].y == p.y){flag = true;break;}}if (!flag){resource.push_back(p);hashChain.push_back(hashTable[n]);hashTable[n] = resource.size() - 1;}}int Serch(point p){int n = Hash(p.x * p.x + p.y * p.y);for (int i = hashTable[n]; i != -1; i = hashChain[i]){if (resource[i].x == p.x && resource[i].y == p.y)return 1;}return 0;}bool op(const point& p1, const point& p2){if (p1.x == p2.x)return p1.y < p2.y;elsereturn p1.x < p2.x;}int main(){while (true){memset(hashTable, -1, sizeof(hashTable));resource.clear();hashChain.clear();int k(0), sum(0);std::cin >> k;if (k == 0) break;std::list<point> temp;for (int i = 0; i < k; ++i){point p;std::cin >> p.x >> p.y;temp.push_back(p);}temp.sort(op);for (std::list<point>::iterator iter = temp.begin(); iter != temp.end(); iter++)Insert(*iter);for (int i = 0; i < k; ++i){for (int j = i + 1; j < k; ++j){if (resource[i].x <= resource[j].x && resource[i].y >= resource[j].y){point p0, p1;p0.x = resource[i].x + resource[j].y - resource[i].y; p0.y = resource[i].y + resource[i].x - resource[j].x;p1.x = resource[j].x + resource[j].y - resource[i].y;p1.y = resource[j].y + resource[i].x - resource[j].x;if (!Serch(p0)) continue;if (!Serch(p1)) continue;sum++;}}}std::cout << sum << std::endl;}}

---