poj1971Parallelogram Counting

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越来越感觉到了数学的重要性！。。

这题本来用以斜率和长度为key值进行hash不过感觉很麻烦还TLE了。。

最后知道中点一样的话就可以组成平行四边形，初中数学就可以了。。

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 #include<map>

11 using namespace std;

12 #define N 1010

13 #define LL long long

14 #define INF 0xfffffff

15 const double eps = 1e-8;

16 const double pi = acos(-1.0);

17 const double inf = ~0u>>2;

18 struct point

19 {

20     int x,y;

21 } p[N],o[N*N];

22 

23 bool cmp(point a,point b)

24 {

25     if(a.x==b.x) return a.y<b.y;

26     return a.x<b.x;

27 }

28 int main()

29 {

30     int t,n,i,j;

31     cin>>t;

32     while(t--)

33     {

34         scanf("%d",&n);

35         for(i = 1 ; i <= n ; i++)

36         {

37             scanf("%d%d",&p[i].x,&p[i].y);

38         }

39         int g = 0;

40         for(i = 1 ; i<= n ; i++)

41         {

42             for(j = i+1 ; j <= n ;j++)

43             {

44                 o[++g].x = (p[i].x+p[j].x);

45                 o[g].y = (p[i].y+p[j].y);

46                // cout<<o[g].y<<endl;

47             }

48         }

49         LL ans = 0;

50         sort(o+1,o+g+1,cmp);

51         int e = 1;

52         for(i = 2; i <= g ;i++)

53         {

54             if(o[i].x-o[i-1].x==0&&o[i].y-o[i-1].y==0)

55             {

56                 e++;

57                // cout<<o[i].x<<" ,"<<o[i-1].x<<" "<<o[i].y<<" "<<o[i-1].y<<endl;

58             }

59             else

60             {

61                 ans+=(LL)(e-1)*e/2;

62                 e = 1;

63             }

64         }

65         ans+=(LL)(e-1)*e/2;

66         cout<<ans<<endl;

67     }

68     return 0;

69 }

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