D - Acute Triangles

思路:

極角排序+點積叉積

在一個三角形中,如果它是直角或者頓角三角形,那麼直角和頓角只會出現一次

所以直角和頓角三角形的個數等於直角和頓角的個數

所以銳角三角形的個數等於三元組個數減去直角和頓角的個數

三點共線看成退化的頓角三角形

怎麼算直角和頓角個數呢, 先按某個點極角排序,然後暴力過取,用雙指標維護到

當前幅角距離為pi/2 到 3*pi/2 的區間, 區間內點的個數就是到當前幅角為直角或頓角的個數

可以用點積和叉積分別判斷角度和相對方向

程式碼:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma
 
 GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-(long double)1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define
 
 pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<long double, long double>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//
 
head

const int N = 2e3 + 5;
struct point {
    LL x, y;
    point(){}
    point(LL x, LL y):x(x), y(y){}
    LL dot(point p) {
        return x*p.x + y*p.y;
    }
    LL cross(point p) {
        return x*p.y - y*p.x;
    }
}p[N], t[N];
bool anglecmp(point a, point b) {
    if(a.y <= 0 && b.y > 0) return true;
    if(a.y > 0 && b.y <= 0) return false;
    if(!a.y && !b.y) return a.x < b.x;
    return a.cross(b) > 0;
}
bool acute(point a, point b) {
    return a.dot(b) > 0 && a.cross(b) >= 0;
}
bool acute2(point a, point b) {
    return a.dot(b) > 0 && a.cross(b) < 0;
}
int main() {
    int T, n;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%lld %lld", &p[i].x, &p[i].y);
        LL ans = 1LL*n*(n-1)*(n-2)/6;
        for (int i = 1; i <= n; i++) {
            int cnt = 0;
            for (int j = 1; j <= n; j++) {
                if(j!=i)
                    t[++cnt] = point(p[j].x - p[i].x, p[j].y - p[i].y);
            }
            sort(t+1, t+1+cnt, anglecmp);
            int l = 1, r = 1;
            for (int j = 1; j <= cnt; j++) {
                while(l <= cnt && acute(t[j], t[l])) l++;
                while(r <= cnt && !acute2(t[j], t[r])) r++;
                ans -= r-l;
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}