【HDU 6731】2019 CCPC秦皇岛 A题（计算几何）

Problem Description

Given n points P1, P2, .... , Pn on 2D plane and q queries. In i-th query, a point Ai is given, and you should determine the number of tuples (u, v) that 1 ≤ u < v ≤ n and Ai , Pu, Pv form a non-degenerate right-angled triangle.

 

 

Input

The first line contains two positive integers n, q (2 ≤ n ≤ 2 000, 1 ≤ q ≤ 2 000), denoting the numberof given points and the number of queries.
Next n lines each contains two integers xi , yi (|xi|, |yi| ≤ 109), denoting a given point Pi.
Next q lines each contains two integers xi , yi (|xi|, |yi| ≤ 109), denoting a query point Ai.
It is guaranteed that the input n + q points are all pairwise distinct.

 

 

Output

Output q lines each contains a non-negative integer, denoting the answer to corresponding query.

 

 

Sample Input

4 2 0 1 1 0 0 -1 -1 0 0 0 1 1

 

 

Sample Output

4 3

 

 

 

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【题目大意】

 给定n个点的x，y，再给定 q 个点，对每个点有一次询问，询问由这个点和前 n 个点能形成多少个直角三角形

【算法分析】

解题思路其实场上已经想的差不多了，但是败在做题少，实现有困难上了（还是做题太少啊，菜哭）

 

分两种情况：

一种是询问点是直角顶点

一种是询问点不是直角顶点

对两种情况，分别枚举，向前 n 个点连边，极角排序

然后再对每个边枚举，直接查看和他垂直的直线是否存在

对第二种情况特殊处理下，只是查询的点略微有些差别

 

对 map 的一些新理解：

只需要重载 < 就能够实现 > 和 == 

 

 

【代码】

 

#include<cstdio>
#include<map>
#include<cstring>
#define ll long long 
using namespace std;
const int MAXN = 2005;
struct P
{
    ll x, y;
    P(ll a=0,ll b=0) 
    {
        x = a, y = b;
    };
    P base()const 
    {
        if (x < 0 || (x == 0 && y < 0))    return P(-x, -y);
        return *this;
    }
    P operator-(const P &a) const 
    {
        return P(x - a.x, y - a.y); 
    }
    bool operator<(const P  &a)const 
    {
        P p1 = base(), p2 = a.base();
        return p1.x*p2.y < p2.x*p1.y;
    }
    
}p[MAXN],q[MAXN];
map<P, int> mapp;
ll ans[MAXN];
int main()
{
    int N,Q;
    while (~scanf("%d%d", &N, &Q))
    {
        memset(ans, 0, sizeof(ans));
        for (int i = 1; i <= N; i++) scanf("%lld %lld", &p[i].x, &p[i].y);
        for (int i = 1; i <= Q; i++) scanf("%lld %lld", &q[i].x, &q[i].y);
        for (int i = 1; i <= Q; i++)
        {
            mapp.clear();
            for (int j = 1; j <= N; j++)
                mapp[(p[j] - q[i])]++;
            for (int j = 1; j <= N; j++)
            {
                P req = p[j] - q[i];
                req = P(-req.y, req.x);
                ans[i] += mapp.count(req) ? mapp[req] : 0;
            }
            ans[i] /= 2;
        }
        for (int i = 1; i <= N; i++)
        {
            mapp.clear();
            for (int j = 1; j <= N; j++)
                if (i != j)    mapp[(p[j] - p[i])]++;
            for (int j = 1; j <= Q; j++)
            {
                P req = q[j] - p[i];
                req = P(-req.y, req.x);
                ans[j] += mapp.count(req) ? mapp[req] : 0;
            }
        }
        for (int i = 1; i <= Q; i++)
        {
            printf("%lld\n", ans[i]);
        }
    }
    return 0;
}