poj2002 数正方形（哈希+几何）

题目大意：给你一堆点，问你能组成几个正方形。

思路：一开始想的是用对角线的长度来当哈希的key，但判断正方形会太复杂，然后就去找了一下正方形的判断方法，发现

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)

就是枚举两个点，然后算出另外两个点。在哈希表中看看能不能找到这两个点。如果只采取其中一个公式的话，切记点的坐标要排序再哈希（具体原因我也不知道，但自己举了很多样例确实是这样），你也可以把两个公式都用上，两个if得到答案。

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<sstream>

#include<cstring>

#include<bitset>

#include<cstdio>

#include<string>

#include<deque>

#include<stack>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#define INF 0x3f3f3f3f

#define CLR(x,y) memset(x,y,sizeof(x))

#define LC(x) (x<<1)

#define RC(x) ((x<<1)+1)

#define MID(x,y) ((x+y)>>1)

using namespace std;

typedef pair<int,int> pii;

typedef long long ll;

const double PI=acos(-1.0);

int fact[10]= {1,1,2,6,24,120,720,5040,40320,362880};

const int maxn= 1010;

const int mod = 100019;

int n;

struct dian {

	int x,y;

} a[maxn];

int hash[mod+10];

int next[maxn];

int gethash(int i) {

	return (a[i].x*a[i].x%mod+a[i].y*a[i].y%mod)%mod;

}

bool find(int x,int y){

	int hashval=(x*x%mod+y*y%mod)%mod;

	for(int i=hash[hashval];i!=-1;i=next[i])

	{

		if(a[i].x==x&&a[i].y==y)return true;

	}

	return false;

}

bool cmp(dian aa,dian bb){

	if(aa.x!=bb.x)

	return aa.x<bb.x;

	return aa.y<bb.y;

}

int main(){

	while(scanf("%d",&n),n)

	{

		int ans=0;

		memset(hash,-1,sizeof(hash));

		for(int i=1;i<=n;i++)

		{

			scanf("%d%d",&a[i].x,&a[i].y);

		}

		sort(a+1,a+1+n,cmp);//非常重要  没有则wa

		for(int i=1;i<=n;i++)

		{

			int hashval=gethash(i);//映射到哈希表里

			next[i]=hash[hashval];

			hash[hashval]=i;

		}

		for(int i=1;i<n;i++)

		{

			for(int j=i+1;j<=n;j++)

			{

				int x=a[i].x-a[j].y+a[i].y;//算出第一个点

				int y=a[i].y+a[j].x-a[i].x;

				if(!find(x,y))continue;//查找

				x=a[j].x-a[j].y+a[i].y;//第二个点

				y=a[j].y+a[j].x-a[i].x;

				if(!find(x,y))continue;//查找

				ans++;

			}

		}

	printf("%d\n",ans/2);//由于有重复计算  所以要除以二

	}

}

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 21208		Accepted: 8136

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.

Sample Input

Sample Output

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poj2002 数正方形（哈希+几何）

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poj2002 数正方形 （哈希+几何）

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