poj 1971 Parallelogram Counting

poj 1971 Parallelogram Counting

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status

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;
}