hdu6731 Angle Beats (ccpc Qinhuangdao A, computational geometry)

Topic links:

http://acm.hdu.edu.cn/showproblem.php?pid=6731

Meaning of the questions:

N-$ $ given points, there are times $ Q $ query

Each time a point is given interrogation $ b $, which seek $ n + 1 $ points, the composition comprises a right triangle $ b $ and how many combinations

data range:

$ 1 \ leq n \ leq $ 2000

$ 1 \ leq q \ leq $ 2000

analysis: 

Category talk

• When the interrogation points as a right angle. Let $ n $ points and $ b $ establish vector, where the vector into the most simple, find mutually perpendicular vectors, vector can be enumerated, seeking vertical vector
• When the initial point is given as a right angle, to the initial point of each offset vector is established, and then enumerate $ b $ points, mutually perpendicular vectors seek

Minimal vector, GCD first removed, if x is negative, the inverse of a vector, if x is 0, y is negative, the inverse of a vector, uniquely determined so that the direction of the vector

Note: int enough, longlong time out

There is also a quicker way: direct deposit x, y does not simplify, when comparing the vector, a direct comparison of the slope, i.e., greater than and less than that if the two vectors are equal

AC Code:

#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
using namespace std;
const int maxn=2007;
pii a[maxn],b[maxn];
int n, q, years [maxn];
int mygcd(int a,int b){
    if(b==0)return a;
    return mygcd(b,a%b);
}
pii getp(int x,int y){
    int tem=mygcd(x,y);
    x / = tem, y / = tem;
    if(x<0)x=-x,y=-y;
    else if(x==0&&y<0)y=-y;
    return make_pair(x,y);
}
map<pii,int>ma;
int main () {
    while(scanf("%d %d",&n,&q)==2){
        for(int i=1;i<=n;i++)scanf("%d %d",&a[i].first,&a[i].second);
        for(int i=1;i<=q;i++)scanf("%d %d",&b[i].first,&b[i].second);
        for(int i=1;i<=q;i++){
            ma.clear();
            for(int j=1;j<=n;j++){
                int x=a[j].first-b[i].first,y=a[j].second-b[i].second;
                ma[getp(x,y)]++;
                if(ma.count(getp(-y,x)))ans[i]+=ma[getp(-y,x)];
            }
        }
      // for(int i=1;i<=q;i++)printf("%d\n",ans[i]),ans[i]=0;
        //cout<<"cdsafa"<<endl;
        for(int i=1;i<=n;i++){
            ma.clear();
            for(int j=1;j<=n;j++){
                if(j==i)continue;
                int x=a[j].first-a[i].first,y=a[j].second-a[i].second;
                ma[getp(x,y)]++;
            }
            for(int j=1;j<=q;j++){
                int x=b[j].first-a[i].first,y=b[j].second-a[i].second;
                if(ma.count(getp(-y,x)))ans[j]+=ma[getp(-y,x)];
            }
        }
        for(int i=1;i<=q;i++)printf("%d\n",ans[i]),ans[i]=0;
    }
    return 0;
}