題解

什麼破題，看一眼就能想出來\(n^2 2^n\)看了一眼資料範圍有點虛，結果跑得飛快= =

處理出\(a[i][j]\)表示從\(i\)到\(j\)經過的點的點集

然後\(f[i][S]\)表示最後一個點在\(i\)處，經過的點集為\(S\)，方案數是多少
然後列舉一個不在\(S\)中的點\(j\)看看\(a[i][j]\)是否全部被\(S\)包含即可

程式碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define pdi pair<db, int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template <class T>
void read(T &res) {
    res = 0;
    char c = getchar();
    T f = 1;
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        res = res * 10 + c - '0';
        c = getchar();
    }
    res *= f;
}
template <class T>
void out(T x) {
    if (x < 0) {
        x = -x;
        putchar('-');
    }
    if (x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 100000007;
int N, pos[(1 << 20) + 5];
int a[25][25];
int f[21][(1 << 20) + 5], cnt[(1 << 20) + 5];
struct Point {
    int x, y;
    Point(int _x = 0, int _y = 0) {
        x = _x;
        y = _y;
    }
    friend Point operator+(const Point &a, const Point &b) { return Point(a.x + b.x, a.y + b.y); }
    friend Point operator-(const Point &a, const Point &b) { return Point(a.x - b.x, a.y - b.y); }
    friend int operator*(const Point &a, const Point &b) { return a.x * b.y - a.y * b.x; }
    friend int dot(const Point &a, const Point &b) { return a.x * b.x + a.y * b.y; }
    int norm() { return x * x + y * y; }
} P[25];
int inc(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
int mul(int a, int b) { return 1LL * a * b % MOD; }
int lowbit(int x) { return x & (-x); }
void update(int &x, int y) { x = inc(x, y); }
void Solve() {
    read(N);
    for (int i = 1; i <= N; ++i) {
        read(P[i].x);
        read(P[i].y);
    }
    for (int i = 1; i <= N; ++i) {
        for (int j = i + 1; j <= N; ++j) {
            a[i][j] |= (1 << i - 1) | (1 << j - 1);
            for (int k = 1; k <= N; ++k) {
                if (k == i || k == j) continue;
                if ((P[k] - P[i]) * (P[j] - P[i]) == 0 && dot(P[k] - P[i], P[j] - P[i]) >= 0 &&
                    (P[k] - P[i]).norm() < (P[j] - P[i]).norm()) {
                    a[i][j] |= (1 << k - 1);
                }
            }
            a[j][i] = a[i][j];
        }
    }
    for (int i = 0; i < N; ++i) pos[1 << i] = i + 1;
    for (int i = 1; i <= N; ++i) {
        f[i][1 << i - 1] = 1;
    }
    for (int S = 1; S < (1 << N); ++S) {
        for (int T = S; T; T -= lowbit(T)) {
            int h = pos[lowbit(T)];
            if (!f[h][S]) continue;
            for (int j = 1; j <= N; ++j) {
                if ((S & (1 << j - 1)) == 0) {
                    if ((a[h][j] & (S | (1 << j - 1))) == a[h][j]) update(f[j][S ^ (1 << j - 1)], f[h][S]);
                }
            }
        }
    }
    int ans = 0;
    for (int S = 1; S < (1 << N); ++S) {
        cnt[S] = cnt[S - lowbit(S)] + 1;
        if (cnt[S] >= 4) {
            for (int i = 1; i <= N; ++i) update(ans, f[i][S]);
        }
    }
    out(ans);
    enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in", "r", stdin);
#endif
    Solve();
    return 0;
}