思路:
极角排序+点积叉积
在一个三角形中,如果它是直角或者顿角三角形,那么直角和顿角只会出现一次
所以直角和顿角三角形的个数等于直角和顿角的个数
所以锐角三角形的个数等于三元组个数减去直角和顿角的个数
三点共线看成退化的顿角三角形
怎么算直角和顿角个数呢, 先按某个点极角排序,然后暴力过取,用双指针维护到
当前幅角距离为pi/2 到 3*pi/2 的区间, 区间内点的个数就是到当前幅角为直角或顿角的个数
可以用点积和叉积分别判断角度和相对方向
代码:
#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; }