Light OJ - 1058 Parallelogram Counting（判定平行四边形）

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

- -

Sample Output
Case :
Case : 6

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

*******************************************************

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[];

 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=,i,j,n;

     scanf("%d", &T);

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

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

         int u=;
         for(i=;i<n-;i++)///保存好向量判断要用到的值
         {
             for(j=i+;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=,f=,sum=;
         for(i=;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-)/;///可以构成平行四边形的数目
                 sum=;
             }
         }

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

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=,i,j,n;

     scanf("%d", &T);

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

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

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

         int u=;
         for(i=;i<n-;i++)
         {
             for(j=i+;j<n;j++)
             {
                 if(p[j].x-p[i].x!=)
                 {
                 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=;
         for(i=;i<u-;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+;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\n",kk++,ans/);///这里显然也天真了
     }
     return ;
 }