K - Parallelogram Counting解题报告（来自网络）

最新推荐文章于 2018-08-09 21:59:00 发布

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最新推荐文章于 2018-08-09 21:59:00 发布

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分类专栏：计算几何专题解题报告文章标签：网络 integer input each 64bit struct

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此博客介绍了一个用于计算平面内给定点构成平行四边形数量的算法，包括输入处理、输出结果以及代码实现。

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K - Parallelogram Counting

Time Limit:5000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1971

Description

There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets of these points that can be written as {A, B, C, D} such that AB || CD, and BC || AD. No four points are in a straight line.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases. It is followed by the input data for each test case.
The first line of each test case contains an integer n (1 <= n <= 1000). Each of the next n lines, contains 2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000.

Output

Output should contain t lines.
Line i contains an integer showing the number of the parallelograms as described above for test case i.

Sample Input

Sample Output

5
6

#include<memory.h>
#include<stdlib.h>
#include<stdio.h>
#define MAX 1000
#define HMAX  0x3fffff
#define P 0x1fffff
//将每个中点的x，y都记录下来，如果
//同时记录拥有这个中点的线段的个数
struct Point{
	int x;
	int y;
	int count;
	int next;
};
Point p[MAX*MAX];
int head[HMAX];
int x[MAX],y[MAX];
int main()
{
	int t,n;
	int zz,xx,yy,xtmp,ytmp,ptr,res,h;
	bool flag;
	scanf("%d",&t);
	while(t--)
	{
		ptr = 1;
		res = 0;
		memset(head,0,sizeof(head));
		memset(p,0,sizeof(p));
		scanf("%d",&n);
		for(int i = 0; i < n; i++)
		{
			scanf("%d%d",&xtmp,&ytmp);
			for(int j = 0; j < i; j++)
			{
				flag = true;
				xx = xtmp + x[j];
				yy = ytmp + y[j];
				//h = xx & P + yy & P; --用这种方法反而更慢...
				//哈希中点坐标之和
				h = abs(xx + yy)%HMAX;
				zz = head[h];
				while(zz){
					if(p[zz].x == xx && p[zz].y == yy){
						p[zz].count++;
						res+=p[zz].count;
						flag = false;
						break;
					}
					else
						zz = p[zz].next;
				}
				//插入新的中点
				if(flag)	{
					p[ptr].x = xx;
					p[ptr].y = yy;
					p[ptr].next = head[h];
					head[h] = ptr++;
				}
			}
			x[i] = xtmp;
			y[i] = ytmp;
		}
		printf("%d/n",res);
	}
	return 0;
}