2017 Russian Code Cup (RCC 17), Elimination Round D

D - Acute Triangles

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
", ans);
    }
    return 0;
}