poj 1971 Parallelogram Counting
2011-12-17 09:45
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Parallelogram Counting
DescriptionThere 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.
InputThe 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.
OutputOutput 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
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 5130 | Accepted: 1699 |
InputThe 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.
OutputOutput should contain t lines.
Line i contains an integer showing the number of the parallelograms as described above for test case i.
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
5 6
#include<iostream> #include<algorithm> using namespace std; struct node { int x; int y; }; node plane[1001]; node middle[500000]; int comp(node a,node b) { if(a.x==b.x) return a.y<b.y; return a.x<b.x; } int main() { int cases; int n; int i,j,k; int sum; int total; int flag; // freopen("E:\\c++\\oj\\t.txt","rt",stdin); cin>>cases; while(cases--) { cin>>n; for(i=0;i<n;i++) { cin>>plane[i].x>>plane[i].y; } k=0; for(i=0;i<n-1;i++) { for(j=i+1;j<n;j++) { middle[k].x=plane[i].x+plane[j].x; middle[k].y=plane[i].y+plane[j].y; k++; } } sort(middle,middle+k,comp); flag=0; total=0; sum=1; for(i=1;i<k;i++) { if(middle[i].x==middle[flag].x && middle[i].y==middle[flag].y) { sum++; } else { flag=i; total+=sum*(sum-1)/2; sum=1; } } cout<<total<<endl; } return 0; }
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