Angle Beats Gym

题目链接：https://vjudge.net/problem/Gym-102361A

题意：给定N个点，q次询问每次询问给一个点，问在N个点中取2个和给定点最多可以组成几个直角三角形。

思路：https://www.cnblogs.com/Jiaaaaaaaqi/p/11631203.html

#include <bits/stdc++.h>
#include <time.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <utility>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <list>
using namespace std;
//cout<<setprecision(10)<<fixed;
#define eps 1e-6
#define PI acos(-1.0)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define mem(s,n) memset(s,n,sizeof s);
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep2(i,a,b) for(int i=a;i>=b;i--)
#define ios {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
typedef long long ll;
typedef unsigned long long ull;
const int maxn=1e6+5;
const ll Inf=0x7f7f7f7f7f7f7f7f;
const ll mod=1e6+3;
//const int N=3e3+5;
bool isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; }//判断一个数是不是 2 的正整数次幂
int modPowerOfTwo(int x, int mod) { return x & (mod - 1); }//对 2 的非负整数次幂取模
int getBit(int a, int b) { return (a >> b) & 1; }// 获取 a 的第 b 位，最低位编号为 0
int Max(int a, int b) { return b & ((a - b) >> 31) | a & (~(a - b) >> 31); }// 如果 a>=b,(a-b)>>31 为 0，否则为 -1
int Min(int a, int b) { return a & ((a - b) >> 31) | b & (~(a - b) >> 31); }
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline int read()
{
    int X=0; bool flag=1; char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();}
    while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
    if(flag) return X;
    return ~(X-1);
}
inline void write(int X)
{
    if(X<0) {X=~(X-1); putchar('-');}
    if(X>9) write(X/10);
    putchar(X%10+'0');
}
/*
inline int write(int X)
{
    if(X<0) {putchar('-'); X=~(X-1);}
    int s[20],top=0;
    while(X) {s[++top]=X%10; X/=10;}
    if(!top) s[++top]=0;
    while(top) putchar(s[top--]+'0');
}
*/
int Abs(int n) {
  return (n ^ (n >> 31)) - (n >> 31);
  /* n>>31 取得 n 的符号，若 n 为正数，n>>31 等于 0，若 n 为负数，n>>31 等于 -1
     若 n 为正数 n^0=n, 数不变，若 n 为负数有 n^(-1)
     需要计算 n 和 -1 的补码，然后进行异或运算，
     结果 n 变号并且为 n 的绝对值减 1，再减去 -1 就是绝对值 */
}
ll binpow(ll a, ll b) {
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = res * a%mod;
    a = a * a%mod;
    b >>= 1;
  }
  return res%mod;
}
void extend_gcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0) {
        x=1,y=0;
        return;
    }
    extend_gcd(b,a%b,x,y);
    ll tmp=x;
    x=y;
    y=tmp-(a/b)*y;
}
ll mod_inverse(ll a,ll m)
{
    ll x,y;
    extend_gcd(a,m,x,y);
    return (m+x%m)%m;
}
ll eulor(ll x)
{
   ll cnt=x;
   ll ma=sqrt(x);
   for(int i=2;i<=ma;i++)
   {
    if(x%i==0) cnt=cnt/i*(i-1);
    while(x%i==0) x/=i;
   }
   if(x>1) cnt=cnt/x*(x-1);
   return cnt;
}
struct node
{
    ll x,y;
    int id;
}p[maxn],be[maxn];
int n,m;
int ans[maxn];
int cmp1(node a,node b)
{
    ll d=a.x*b.y-b.x*a.y;
    if(d==0) return a.x<b.x;
    else return d>0;
}
int qua(node a)
{
    if(a.x>0&&a.y>=0) return 1;
    else if(a.x<=0&&a.y>0) return 2;
    else if(a.x<0&&a.y<=0) return 3;
    else if(a.x>=0&&a.y<=0) return 4;
}
int cmp(node a,node b)
{
    if(qua(a)==qua(b)) return cmp1(a,b);
    else return qua(a)<qua(b);
}
ll check(node a,node b)
{
    return a.x*b.x+a.y*b.y;
}
ll chaji(node a,node b)
{
    return a.x*b.y-b.x*a.y;
}
ll work(node pp)
{
    for(int i=1;i<=n;i++)
    {
        p[i]=be[i];
        p[i].x-=pp.x;
        p[i].y-=pp.y;
    }
    p[0]=pp;
    sort(p+1,p+n+1,cmp);
    for(int i=1;i<=n;i++)
    {
        p[i+n]=p[i];
    }
    ll ans=0;
    int r=2;
    for(int l=1;l<=n;l++)
    {
        while(r<=2*n)
        {
            if(chaji(p[l],p[r])<0) break;
            if(check(p[l],p[r])<=0) break;
            r++;
        }
        int tr=r;
        while(tr<=2*n)
        {
            if(chaji(p[l],p[tr])<=0) break;
            if(check(p[l],p[tr])!=0) break;
            ans++;
            tr++;
        }
    }
    return ans;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int all=0;
        for(int i=1;i<=n;i++)
        {
            all++;
            int x,y;
            scanf("%d%d",&x,&y);
            p[all].x=x,p[all].y=y,p[all].id=0;
            be[all]=p[all];
        }
        for(int i=1;i<=m;i++)
        {
            all++;
            int x,y;
            scanf("%d%d",&x,&y);
            p[all].x=x,p[all].y=y,p[all].id=i;
            be[all]=p[all];
            ans[i]=work(p[all]);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=all;j++)
            {
                p[j]=be[j];
            }
            p[0]=be[i];
            int flag=0;
            for(int j=1;j<=all;j++)
            {
                if(p[j].x==p[0].x&&p[j].y==p[0].y) flag=1;
                if(flag) p[j]=p[j+1];
                p[j].x-=p[0].x;
                p[j].y-=p[0].y;
            }
            int nn=all-1;
            sort(p+1,p+1+nn,cmp);
            for(int j=1;j<=nn;j++)
            {
                p[j+nn]=p[j];
            }
            int r=2;
            for(int l=1;l<=nn;l++)
            {
                int id=0;
                if(p[0].id) id=p[0].id;
                if(p[l].id) id=p[l].id;
                while(r<=2*nn)
                {
                    if(chaji(p[l],p[r])<0) break;
                    if(check(p[l],p[r])<=0) break;
                    r++;
                }
                int tr=r;
                while(tr<=2*nn)
                {
                    if(chaji(p[l],p[tr])<=0) break;
                    if(check(p[l],p[tr])!=0) break;
                    if(id==0)
                    {
                        if(p[tr].id) ans[p[tr].id]++;
                    }
                    else
                    {
                        if(p[tr].id==0) ans[id]++;
                    }
                    tr++;
                }
            }
        }
        for(int i=1;i<=m;i++)
        {
            printf("%d
",ans[i]);
        }
    }
    return 0;
}