topic

Description

　　A two-dimensional plane coordinate system has N points.
　　3 points selected from N points, asked how many such method is selected from these three points form a right triangle.

Input

　　The first row contains an integer N (3 <= N <= 1500), number of dots.
　　Next N rows, each row comprising two integers separated by a space coordinates of each point represent the coordinate values between -10 ^ 9 to 10 ^ 9.
　　Each point different from each other.

Output

　　Output of the number of right-angled triangle.

Sample Input

Input 1: 
3 
. 4 2 
2 1 
1 3 

input 2: 
. 4 
. 5 0 
2. 6 
. 8. 6 
. 5. 7 

Input 3: 
. 5 
-1 1 
-1 0 
0 0 
1 0 
1 1

Sample Output

Output 1: 
1 

Output 2: 
0 

Output 3: 
7

Data Constraint

analysis

First of all, we certainly want to enumerate twenty-two
Then, we need to know the slope
In the coordinate system a linear function y = kx k is the slope
Quadratic function is y = kx + bk slope
Because both sides of a right triangle slope multiplied by -1
With this property, we can see that the slope of the triangular configuration but equal in the different quadrants of point
So we ask the same slope point
Then we can have a sort to optimize
With the same constant slope
The answer is to speak the same slope adjacent two quadrants of the dot and then summed up

Code

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 long long x[10010],y[10001];
 5 struct sb
 6 {
 7     long long x,y,xx;
 8 }tmp[10001];
 9 void turn(int p)
10 {
11     swap(tmp[p].x,tmp[p].y);
12     tmp[p].y=-tmp[p].y;
13     tmp[p].xx=(tmp[p].xx+1)%4;
14 }
15 bool cmp(sb a,sb b)
16 {
17     return a.x*b.y>b.x*a.y;
18 }
19 int cnt;
20 int main ()
21 {
22     int n;
23     cin>>n;
24     for (int i=1;i<=n;i++) cin>>x[i]>>y[i];
25     for (int i=1;i<=n;i++)
26     {
27         for (int j=1;j<=n;j++)
28         {
29             tmp[j].x=x[j]-x[i];
30             tmp[j].y=y[j]-y[i];
31             tmp[j].xx=0;
32             if (i==j)
33             {
34                 tmp[j].x=tmp[1].x;
35                 tmp[j].y=tmp[1].y;
36                 tmp[j].xx=tmp[1].xx;
37             }
38             else 
39               while (tmp[j].x<=0||tmp[j].y<0) turn(j);
40         }
41         sort(tmp+2,tmp+n+1,cmp);
42         int j=2;
43         while (j<=n)
44         {
45             int ans[4]={0};
46             int k=j;
47             while (k<=n&&tmp[j].x*tmp[k].y==tmp[k].x*tmp[j].y)
48                 ans[tmp[k].xx]++,k++;
49             for (int kk=0; kk <= 3 ; kk ++ )
 50                cnt + = years [kk] * years [(kk + 1 )% 4 ];
51              j = k; 
52          }
 53      }
 54      cout << cnt;
55 }

JZOJ 1385. right triangle