1058 - Parallelogram Counting
2016-08-15 11:39
435 查看
1058 - Parallelogram Counting
PDF (English) Statistics Forum
Time Limit: 2 second(s) Memory Limit: 32 MB
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
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
平行四边形的判定定理:
两组对边分别平行的四边形是平行四边形(定义判定法);
一组对边平行且相等的四边形是平行四边形;
两组对边分别相等的四边形是平行四边形;
两组对角分别相等的四边形是平行四边形(两组对边平行判定);
对角线互相平分的四边形是平行四边形。
这道题用定理5判断。
记录每条边的中点坐标,如果两个边的中点坐标相同,证明这两条边为一个平行四边形的两条对角线。
PDF (English) Statistics Forum
Time Limit: 2 second(s) Memory Limit: 32 MB
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
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
平行四边形的判定定理:
两组对边分别平行的四边形是平行四边形(定义判定法);
一组对边平行且相等的四边形是平行四边形;
两组对边分别相等的四边形是平行四边形;
两组对角分别相等的四边形是平行四边形(两组对边平行判定);
对角线互相平分的四边形是平行四边形。
这道题用定理5判断。
记录每条边的中点坐标,如果两个边的中点坐标相同,证明这两条边为一个平行四边形的两条对角线。
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <queue> #include <stack> using namespace std; struct node { int x;int y; }; node di[1010]; node pare[1000010]; int n; bool cmp(node a,node b){ if(a.x==b.x) return a.y<b.y; return a.x<b.x; } int main(){ int T,i,j,t=1; scanf("%d",&T); while(T--){ scanf("%d",&n); for( i=1; i<=n;++i ) scanf("%d %d",&di[i].x,&di[i].y); int k=0; for( i=1;i<=n;i++){ for( j=i+1;j<=n;j++){ pare[++k].x=di[i].x+di[j].x; pare[k].y=di[i].y+di[j].y; } } sort(pare+1,pare+k+1,cmp); int temp=1; int ans=0; for( i=2;i<=k;i++){ if(pare[i].x==pare[i-1].x&&pare[i].y==pare[i-1].y) temp++; else{ ans+=temp*(temp-1)/2; temp=1; } } printf("Case %d: %d\n",t++,ans); } return 0; }
相关文章推荐
- lightoj1058 Parallelogram Counting
- LightOJ 1058 - Parallelogram Counting
- Light OJ:1058Parallelogram Counting(几何数学+技巧排序)
- LightOJ 1058 & Poj 1971 Parallelogram Counting
- 【LightOJ - 1058】Parallelogram Counting
- lightoj-1058 - Parallelogram Counting
- POJ - 1971 - Parallelogram Counting
- POJ 1971 Parallelogram Counting
- POJ 1972 Parallelogram Counting
- poj 1971 Parallelogram Counting
- poj 1971 Parallelogram Counting
- POJ 1971 Parallelogram Counting
- poj1971:Parallelogram Counting
- poj 1971 Parallelogram Counting
- POJ 1971 Parallelogram Counting
- poj 1971 Parallelogram Counting
- Parallelogram Counting
- d-Left Counting Bloom Filter (5)
- 计数排序源码(counting sort)
- HDOJ 1058 Humble Numbers [简单DP]