pku 2002 Squares 二分查找 或者 hash

http://poj.org/problem?id=2002

首先枚举正方形的任意两点，然后有由这两点确定其他两点，确定后再到给定的点中做二分查找如果都存在，则找到了一个正方形。首先排序，然后查找，记住这里重复计算了，要除以2

推导过程转载自：http://blog.csdn.net/zsc09_leaf/article/details/6204705

公式推导 :
//  如图所示,正方形的四个点分别为 (x1,y1),(x2,y2),(x3,y3),(x4,y4)
//  假设一开始给出的点为  (x1,y1),(x2,y2) 且已排序
//  根据全等三角形,可以得知2个蓝色△全等，2个红色△全等 (注:此处也可推右边的正方形)
//  所以可以推出 x3=x2-(y2-y1);
//               y3=y2+(x2-x1);
//               x4=x1-(y2-y1);
//               y4=y1+(x2-y1);
//  当然全体是排好序。 因为这样子做会导致边被重用，所以需除于 2 ，具体追究可自己详细思考

#include <iostream>
#include <cstdio>
#include <algorithm>
#define maxn 1007
using namespace std;

struct node
{
    int x,y;
}p[maxn];
int n;

int cmp(node a,node b)
{
    if (a.x == b.x) return a.y <b.y;
    else   return a.x < b.x;
}
bool bsearch(int x,int y,int l,int r)
{
    while (l <= r)
    {
        int m = (l + r)/2;
        if (p[m].x == x && p[m].y == y) return true;
        else if (p[m].x > x)
        {
            r = m - 1;
        }
        else if (p[m].x < x)
        {
            l = m + 1;
        }
        else
        {
            if (p[m].y > y)
            r = m -1;
            else
            l = m + 1;
        }
    }
    return false;
}
int main()
{
    int i,j,ans;
    while (~scanf("%d",&n))
    {
        if (!n) break; ans = 0;
        for (i = 0; i < n; ++i)
        scanf("%d%d",&p[i].x,&p[i].y);
        sort(p, p + n,cmp);
        for (i = 0; i < n; ++i)
        {
            for (j = i + 1; j < n; ++j)
            {
                int x3=p[j].x-(p[j].y-p[i].y);
                int y3=p[j].y+(p[j].x-p[i].x);
                if(!bsearch(x3,y3,0,n))
                    continue;
                int x4=p[i].x-(p[j].y-p[i].y);
                int y4=p[i].y+(p[j].x-p[i].x);
                if(!bsearch(x4,y4,0,n))
                    continue;
                ans++;
            }
        }
        printf("%d\n",ans/2);
    }
    return 0;
}

hash开散列挂链实现：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 1007
#define N 9997
using namespace std;

struct mode
{
    int x,y;
}p[maxn];
struct node
{
    int x,y;
    node *next;
}*pp[N],H[N + 25];
int n,pos;

int cmp(mode a,mode b)
{
    if (a.x == b.x) return a.y <b.y;
    else   return a.x < b.x;
}
bool bsearch(int x,int y)
{
    int tmp = (x + y)%N;
    if (tmp < 0) tmp = -tmp;
    node *q;
    for (q = pp[tmp]; q != NULL; q = q->next)
    {
        if (q->x == x && q->y == y)
        return true;
    }
    return false;
}
int main()
{
    int i,j,ans;
    while (~scanf("%d",&n))
    {
        if (!n) break; ans = pos = 0;
        memset(pp,0,sizeof(pp));
        for (i = 0; i < n; ++i)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
            int tmp = p[i].x + p[i].y;
            int l = tmp%N;
            if (l < 0) l = -l;
            node *t = &H[pos++];
            t->x = p[i].x;
            t->y = p[i].y;
            t->next = pp[l];
            pp[l] = t;
        }
        sort(p, p + n,cmp);
        for (i = 0; i < n; ++i)
        {
            for (j = i + 1; j < n; ++j)
            {
                int x3=p[j].x-(p[j].y-p[i].y);
                int y3=p[j].y+(p[j].x-p[i].x);
                if(!bsearch(x3,y3))
                    continue;
                int x4=p[i].x-(p[j].y-p[i].y);
                int y4=p[i].y+(p[j].x-p[i].x);
                if(!bsearch(x4,y4))
                    continue;
                ans++;
            }
        }
        printf("%d\n",ans/2);
    }
    return 0;
}