Qinhuangdao station 2019CCPC A.Angel Beats

That Italy: the plane of the points given n, q interrogation times, to a given time point P, Q n points in this plane and the point number right triangle can be composed, wherein (n + q) points equal to each other

Ideas:

Consider a point P at a right angle, respectively for non-right angle vertex and vertices. This question is very simple idea is to see how simple and does not tle! ! !

For the right-angled vertex and non-right angle vertex codes are relatively simple, which is a little off the idea of ​​seeking.

Here is a map of the usage of say, less than custom operators, so that when looking at the map, the slope of the vector will be unified together, although a number of different vectors will still be stored in the map. (I heard that writing a scene of blood, Zhongshan University Gangster nb)

Algorithm complexity about n * n * log (n) (Due to the same range of n and p, where a uniform representation with n), the running time of about 10s

Code:

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 2005;
struct P
{
    ll x, y;
    P(ll xx=0, ll yy=0) {
        x = xx; y = yy;
    }
    P base()const{
        if (x < 0 || (x == 0 && y < 0))return P(-x, -y);
        return *this;
    }
    BOOL  operator <( const P & B) const { 
        P P1 = Base ();. P2 = P B Base ();
        // if collinear, are considered the same index 
        return P1.x p2.y * <p1.y * P2 .x; 
    } 
    P operator - ( const P & B) const {
         return P (X - BX, Y - by); 
    } 
} A [N], Qur'an [N]; 
int n-, Q; 
Map <P, int > m; 
ANS LL [N]; 
int main () {
     the while (~ Scanf ( "%d%d", &n, &q)) {
        memset(ans,0,sizeof(ans));
        for (int i = 0; i < n; i++)scanf("%lld%lld", &a[i].x, &a[i].y);
        for (int i = 0; i < q; i++)scanf("%lld%lld", &qur[i].x, &qur[i].y);
        for (int i = 0; i < q; i++) {
          //求解作为直角顶点
            m.clear();
            for( Int J = 0 ; J <n-; J ++ ) 
                m [A [J] - Qur'an [I]] ++ ;
             for ( int J = 0 ; J <n-; J ++ ) { 
                P P = A [J] - Qur'an [I]; 
                P = P (- Py, PX); 
                ANS [I] + = m.count (P) m [P]:? 0 ; 
            } 
　　　　　　　// Since the two right-angle sides are enumerated, so in addition to 2 
            ANS [I] / = 2 ; 
        } 
        for ( int I = 0 ; I <n-; I ++ ) {
            //As a non-right-angled vertex, each enumeration point i, as a right-angled vertex, all of the updated set of q interrogation points 
            m.clear ();
             for ( int J = 0 ; J <n-; J ++ ) {
                 IF (! I = J) m [A [J] - A [I]] ++ ; 
            } 
            for ( int J = 0 ; J <Q; J ++ ) { 
                P P = Qur'an [J] - A [I]; 
                P = P (- Py, PX); 
                ANS [J] + = m.count (P) m [P]:? 0 ; 
            } 
        } 
        for ( int I = 0 ; I <Q; I ++) the printf ("%lld\n", ans[i]);
    }
     return 0;
}