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, contains 2 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
找所有的连线的中点。相同的就加一下，因为两条中点相同的直线可以构成平行四边形。
//数平行四边形？ #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int maxn = 1010; struct node{ int x,y; }; node mid[maxn*maxn],s[maxn]; bool cmp(node a,node b){ return a.x == b.x?a.y<b.y:a.x<b.x; } int main(){ int t,n,text = 1; int next,sum,cnt; scanf("%d",&t); while(t--){ scanf("%d",&n); for(int i=0; i<n; i++){ scanf("%d%d",&s[i].x,&s[i].y); } int top = 0; for(int i=0; i<n; i++){ for(int j=i+1; j<n; j++){ mid[top].x = s[i].x+s[j].x; mid[top++].y = s[i].y+s[j].y; } } sort(mid,mid+top,cmp); next = 0;sum = 1;cnt = 0; for(int i=1; i<top; i++){ if(mid[i].x == mid[next].x && mid[i].y == mid[next].y){ sum++; } else{ next = i; cnt+=sum*(sum-1)/2; sum = 1; } } printf("Case %d: %d\n",text++,cnt); } return 0; }
