LightOJ - 1058 Parallelogram Counting （數學幾何&技巧）給n個點求組成平行四邊形個數

LightOJ - 1058 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 //題意：輸入一個n，再輸入n個點的坐标。 給你n個點的坐标讓你求出這n個點可以組成幾個平行四邊形。 //思路： 因為平行四邊形的兩條對角線的交點是唯一的，是以先求出這n個點所能組成的所有線段的中點（n*(n-1)/2個中點），對其進行排序後，對這些中點進行計算，如果在一個中點處有num條線段相交，那麼在這個中點處可以組成num*(num-1)/2個平行四邊形。最所有中點模拟一遍對其求和即為所得。 #include<stdio.h> #include<string.h> #include<algorithm> #define INF 0x3f3f3f3f #define ll long long #define N 1010 #define M 1000000007 using namespace std; struct zz { int x; int y; }p[N],mid[N*N]; int cmp(zz a, zz b) { if(a.x==b.x) return a.y<b.y; return a.x<b.x; } int main() { int t,n,i,j,T=1; scanf("%d",&t); while(t--) { scanf("%d",&n); for(i=0;i<n;i++) scanf("%d%d",&p[i].x,&p[i].y); //sort(p,p+n,cmp); int k=0; for(i=0;i<n-1;i++) { for(j=i+1;j<n;j++) { mid[k].x=p[j].x+p[i].x; mid[k++].y=p[j].y+p[i].y; } } sort(mid,mid+k,cmp); // for(i=0;i<k;i++) // printf("%d %d-----\n",mid[i].x,mid[i].y); // printf("\n"); int num=1; int sum=0; for(i=0;i<k-1;i++) { if(mid[i].x==mid[i+1].x&&mid[i].y==mid[i+1].y) num++; else { sum+=(num*(num-1)/2); num=1; } } printf("Case %d: %d\n",T++,sum); } return 0; }