Light OJ

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

Input starts with an integer T (≤ 15), denoting the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 1000). Each of the next n lines, contains2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000.

Output

For each case, print the case number and the number of parallelograms that can be formed.

Sample Input
2
6
0 0
2 0
4 0
1 1
3 1
5 1
7
-2 -1
8 9
5 7
1 1
4 8
2 0
9 8

Sample Output
Case 1: 5
Case 2: 6

题目链接：http://lightoj.com/volume_showproblem.php?problem=1058

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题意：给出T组实例，每组实例给出n各个点的坐标，判断能后成多少个平行四边形。

分析：使用向量法判断，只要满足x1+x3=x2+x4,y1+y3=y2+y4，则说明可以构成平行四边行

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <map>
#include <vector>
using namespace std;

#define N 1200

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

node q[500000];

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

int main()
{
    int T ,kk=1,i,j,n;

    scanf("%d", &T);

    while(T--)
    {
        scanf("%d", &n);

        for(i=0;i<n;i++)
            scanf("%d%d", &p[i].x,&p[i].y);

        int u=0;
        for(i=0;i<n-1;i++)///保存好向量判断要用到的值
        {
            for(j=i+1;j<n;j++)
            {
                q[u].x=p[i].x+p[j].x;
                q[u++].y=p[i].y+p[j].y;
            }
        }

        sort(q,q+u,cmp);

        int ans=0,f=0,sum=1;
        for(i=1;i<u;i++)///使用向量法判定是否满足为平行四边形
        {
            if(q[i].x==q[f].x&&q[i].y==q[f].y)
                sum++;
            else
            {
                f=i;///
                ans+=sum*(sum-1)/2;///可以构成平行四边形的数目
                sum=1;
            }
        }

        printf("Case %d: %d
",kk++,ans);
    }
    return 0;
}

附上一个RE的代码：（使用的斜率和b值判断的）指出一些错误和注意事项

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <map>
#include <vector>
using namespace std;

#define N 12000
#define INF 0x3f3f3f3f

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

struct no
{
    double k,l,b;
}q[N];///应为q[500000]，对于储存统计比较条件的这个数组一定要开的更大，依旧为N的话RE无止境。。。

double Len(node a,node b)
{
    double x=b.x-a.x;
    double y=b.y-a.y;
    double len=sqrt(x*x+y*y);
    return len;
}

int main()
{
    int T ,kk=1,i,j,n;

    scanf("%d", &T);

    while(T--)
    {
        scanf("%d", &n);

        memset(p,0,sizeof(p));
        memset(q,0,sizeof(q));

        for(i=0;i<n;i++)
            scanf("%d%d", &p[i].x,&p[i].y);

        int u=0;
        for(i=0;i<n-1;i++)
        {
            for(j=i+1;j<n;j++)
            {
                if(p[j].x-p[i].x!=0)
                {
                q[u].k=1.0*(p[j].y-p[i].y)/(p[j].x-p[i].x);
                q[u].b=p[i].y-q[u].k*p[i].x;
                q[u++].l=Len(p[i],p[j]);
                }
                else
                {
                    q[u].k=INF;
                    q[u].b=p[i].x;
                    q[u++].l=Len(p[i],p[j]);
                }
            }
        }

        int ans=0;
        for(i=0;i<u-1;i++)///上面的N改了之后就知道这双重for提交，TLE妥妥的，所以这里思路必须得换一下
        {
            for(j=i+1;j<u;j++)///然后我就直接改方法了，也不知道这个路子能被改出正确答案不%>_<%
            {
                    if((q[i].k==q[j].k)&&(q[i].l==q[j].l)&&q[i].b != q[j].b)
                    ans++;
            }
        }

        printf("Case %d: %d
",kk++,ans/2);///这里显然也天真了
    }
    return 0;
}