POJ 2002 Squares 计算集合 点的hash

题目大意：给出平面上的n个点，问能组成多少个正方形。

思路：一开始看时间3秒半，就想用set水过，然而失败了。没办法手写hash吧。观察坐标的范围，<20000，样例中还有负的，我们读进来的时候就将点的坐标+20000，这样避免负数，方便hash。我的哈希很弱，就是把xy坐标加起来作为哈希值，想卡的话应该很轻松。但还是过得很快。

CODE：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 1010using namespace std;struct Point{int x,y;Point(int _ = 0,int __ = 0):x(_),y(__) {}bool operator ==(const Point &a)const {return (x == a.x && y == a.y);}Point operator -(const Point &a)const {return Point(x - a.x,y - a.y);}Point operator +(const Point &a)const {return Point(x + a.x,y + a.y);}void Read() {scanf("%d%d",&x,&y);x += 20000,y += 20000;}}point[MAX];struct HashSet{int head[80010],total;int next[MAX];Point val[MAX];void Insert(const Point &p) {int x = p.x + p.y;next[++total] = head[x];val[total] = p;head[x] = total;}bool Find(const Point &p) {int x = p.x + p.y;for(int i = head[x]; i; i = next[i])if(val[i] == p)return true;return false;}void Clear() {memset(head,0,sizeof(head));total = 0;}}hash;int points;int main(){while(scanf("%d",&points),points) {for(int i = 1; i <= points; ++i) {point[i].Read();hash.Insert(point[i]);}int ans = 0;for(int i = 1; i <= points; ++i)for(int j = i + 1; j <= points; ++j) {Point v = point[j] - point[i];Point _v(v.y,-v.x);if(hash.Find(point[i] + _v) && hash.Find(point[j] + _v))++ans;Point __v(-v.y,v.x);if(hash.Find(point[i] + __v) && hash.Find(point[j] + __v))++ans;}printf("%d\n",ans / 4);hash.Clear();}return 0;}