POJ2002 Squares（hash）

给出N个点，问这些点能组成多少个正方形，N小于1000，显然的，思路是，先确定正方形的两个顶点，然后在点集中查找另外两个顶点

至于查找，有两种做法，一个是二分，一个是hash，很明显，hash要快一些

Node 1：二分 1157ms，具体见代码

#include <iostream>
#include <algorithm>
#include <cstdio>

using namespace std;
struct node
{
	int x, y;
} p[1001];
bool op(node xx,node yy)
{
	if (xx.x == yy.x)
		return xx.y < yy.y;
	else return xx.x < yy.x;
}
int main()
{
	int points;
	while (scanf("%d", &points), points)
	{
		int sum = 0;
		for (int k = 0; k < points; ++k)
			scanf("%d%d", &p[k].x, &p[k].y);
		sort(p, p + points, op);
		for (int i = 0; i < points; ++i)
		{
			for (int j = i + 1; j < points; ++j)
			{
				if(p[i].x <= p[j].x && p[i].y >= p[j].y)
				{
				node p0, p1;
				p0.x = p[i].x + p[j].y - p[i].y;
				p0.y = p[i].y + p[i].x - p[j].x;
				p1.x = p[j].x + p[j].y - p[i].y;
				p1.y = p[j].y + p[i].x - p[j].x;
				if (!binary_search(p, p+points, p0, op))
					continue;
				if (!binary_search(p, p+points, p1, op))
					continue;
				sum++;
				}
			}
		}
		printf("%d\n", sum);
	}
	return 0;
}

Node 2：hash，735ms

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int N = 1010;
const int Mod = 11117;

int ans;

struct Point{
	int x,y;
}p[N];

struct Node{
	Point x;
	Node *next;
}node[Mod+10];

bool cmp(Point a,Point b){
	if(a.x==b.x)return a.y<b.y;
	return a.x<b.x;
}

void insert(Point a){  
    int key=(a.x*a.x+a.y*a.y)%Mod;
	Node *newNode = new Node;  
    newNode->x=a; 
	newNode->next = node[key].next;
	node[key].next = newNode;  
}  

int search(Point a){  
    int key=(a.x*a.x+a.y*a.y)%Mod;
	Node *shit = &node[key];
	while(shit != NULL){    
        if( shit->x.x == a.x && shit->x.y == a.y)return 1;
		shit = shit->next;
    }
	return 0;
}

int main()
{
	int i,j,n;
	while(scanf("%d",&n),n){ 
        memset(node,0,sizeof(node));
		for(i=0;i<n;i++){
			scanf("%d%d",&p[i].x,&p[i].y);
			insert(p[i]);
		}
		//开始没有排序，下面只是查找了p0，p1两个点，一直wa，因为不排序的话，需要查找4个点，然后结果除以4
		sort(p,p+n,cmp);
		ans=0;
		for (i = 0; i < n; ++i){
			for (j = i + 1; j < n; ++j){
					Point p0,p1;
					p0.x = p[i].x + p[i].y - p[j].y;  
					p0.y = p[i].y - p[i].x + p[j].x;
					if( !search(p0) )continue;

					p1.x = p[j].x + p[i].y - p[j].y;  
					p1.y = p[j].y - p[i].x + p[j].x;
					if( !search(p1) )continue;
					ans++;
			}
		}
		printf("%d\n",ans/2);	
	}
	return 0;
}