HDU - 5784 How Many Triangles 极角排序 + 旋转卡壳

HDU - 5784

感觉好久没写这种题了啊。。感觉细节好多。

对每个点极角排序统计钝角和直角， 还有三点共线的。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0)                     ;

using namespace std;

const int N = 4000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

struct Point {
    LL x, y;
    Point(LL x = 0, LL y = 0) : x(x), y(y) {}
    Point operator + (const Point &rhs) const {
        return Point(x + rhs.x, y + rhs.y);
    }
    Point operator - (const Point &rhs) const {
        return Point(x - rhs.x, y - rhs.y);
    }
    int quan() {
        return y < 0 || y == 0 && x < 0;
    }
};


LL Det(const Point &A, const Point &B) {
    return A.x * B.y - B.x * A.y;
}
LL Dot(const Point &A, const Point &B) {
    return A.x * B.x + A.y * B.y;
}

bool operator < (Point A, Point B) {
    if(A.quan() != B.quan()) return A.quan() < B.quan();
    else return Det(A, B) > 0;
}

bool equ(Point A, Point B) {
    if(A < B) return false;
    if(B < A) return false;
    return true;
}

LL c2(int x) {
    return 1LL * x * (x - 1) / 2;
}

int n;
Point a[N], b[N];

bool check(Point A, Point B) {
    if(Det(A, B) == 0) return false;
    if(Dot(A, B) <= 0) return false;
    return true;
}

int main() {
    while(scanf("%d", &n) != EOF) {
        for(int i = 1; i <= n; i++) {
            scanf("%lld%lld", &a[i].x, &a[i].y);
        }
        LL ans = 0, tmp = 0;
        for(int o = 1; o <= n; o++) {
            int tot = 0;
            for(int j = 1; j <= n; j++) {
                if(j == o) continue;
                b[tot++] = a[j] - a[o];
            }
            sort(b, b + tot);
            for(int i = 0; i < tot; i++) {
                b[i + tot] = b[i];
            }
            int pt1 = 0, pt2 = 0;
            for(int i = 0, j = 0; i < tot; i = j) {
                while(j < i + tot && equ(b[i], b[j])) j++;
                chkmax(pt1, j);
                while(pt1 < i + tot && Dot(b[i], b[pt1]) > 0 && Det(b[i], b[pt1]) > 0) pt1++;
                chkmax(pt2, pt1);
                while(pt2 < i + tot && Det(b[i], b[pt2]) > 0) pt2++;
                ans += 1LL * (pt2 - pt1) * (j - i);
            }
            for(int i = 0, j = 0; i < tot; i = j) {
                j = i;
                while(j < i + tot && equ(b[i], b[j])) j++;
                tmp += c2(j - i);
            }
        }
        ans = 1LL * n * (n - 1) * (n - 2) / 6 - ans - tmp / 2;
        printf("%lld\n", ans);
    }
    return 0;
}

/*
*/