LightOJ 1058 - Parallelogram Counting
2016-08-14 20:38
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1058 - Parallelogram Counting
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.
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 than1000000000.
SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET)
平行四边形的性质,对角线互相平分,,求任意两个点的中点。找中点一样的线段,任意取两条就是平行四边形了。
PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 than1000000000.
Output
For each case, print the case number and the number of parallelograms that can be formed.Sample Input | Output for 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 | Case 1: 5 Case 2: 6 |
SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET)
平行四边形的性质,对角线互相平分,,求任意两个点的中点。找中点一样的线段,任意取两条就是平行四边形了。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct node { int x,y; }t[1100],t1[550000]; bool cmp(node a,node b) { if(a.x!=b.x) return a.x<b.x; return a.y<b.y; } int main() { int T,n,i,j; scanf("%d",&T); int cnt=1; while(T--) { scanf("%d",&n); for(i=0;i<n;i++) scanf("%d%d",&t[i].x,&t[i].y); int k=0; for(i=0;i<n-1;i++) for(j=i+1;j<n;j++) { t1[k].x=t[i].x+t[j].x; t1[k].y=t[i].y+t[j].y; k++; } sort(t1,t1+k,cmp); int l=1; int sum=0; for(i=1;i<k;i++) { if(t1[i].x==t1[i-1].x&&t1[i].y==t1[i-1].y) { l++; } else { sum+=l*(l-1)/2; l=1; } } printf("Case %d: %d\n",cnt++,sum); } return 0; }
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