题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=5020

https://loj.ac/problem/2289

题解

这个 appear 和 disappear 操作显然是强行加上去用力啊增加代码长度的。

所以相当于就是什么东西套个 LCT 就行了。

所以考虑怎么快速求出一堆东西的分值和。

\(sin, exp\),一次函数之间的加法似乎并没有什么优美的性质,所以我们考虑泰勒展开。
\[e^v = \sum_{i=0}^{\infty} \frac1{i!} \cdot v^i\]
我们把 \(v=ax+b\) 带进去,就是
\[\begin{align*}e^v &= \sum_{i=0}^{\infty} \frac1{i!} \cdot (ax+b)^i\\&= \sum_{i=0}^{\infty} \frac1{i!} \sum_{j=0}^i \binom ij a^jb^{i-j}x^j\\&= \sum_{j=0}^{\infty}a^jx^j \sum_{i=j}^{\infty} \frac{\binom ij b^{i-j}}{i!}\end{align*}\]
这样我们就可以求出每一个 \(x^i\) 前面的系数了。

\(sin\) 的话同理,就不重新写一遍了。一次函数的话 \(x^i\) 前面的系数根本不用算。

大概展开十几项就够了,这里开了 \(16\) 项。


这样话时间复杂度就是 \(O(m(16\log n+16^2))\)

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
    x = c & 15;
    while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
    f ? x = -x : 0;
}

const int N = 100000 + 7;
const int M = 16;

#define lc c[0]
#define rc c[1]

int n, m;
ll fac[M], C[M][M];

struct Node {
    int c[2], fa, rev;
    double v[M], sum[M];
    inline Node() {}
    inline void set(const int &f, const double &a, const double &b) {
        memset(v, 0, sizeof(v));
        if (f == 1) {
            double aa = 1;
            for (int i = 0; i < M; ++i) {
                double bb = 1;
                for (int j = i; j < M; ++j, bb *= b)
                    if ((j & 1) && (j >> 1) & 1) v[i] -= C[j][i] * bb / fac[j];
                    else if ((j & 1) && !((j >> 1) & 1)) v[i] += C[j][i] * bb / fac[j];
                v[i] *= aa, aa *= a;
            }
        } else if (f == 2) {
            double aa = 1;
            for (int i = 0; i < M; ++i) {
                double bb = 1;
                for (int j = i; j < M; ++j, bb *= b) v[i] += C[j][i] * bb / fac[j];
                v[i] *= aa, aa *= a;
            }
        } else if (f == 3) v[0] = b, v[1] = a;
    }
} t[N];
int st[N];
inline bool idtfy(int o) { return t[t[o].fa].rc == o; }
inline bool isroot(int o) { return t[t[o].fa].lc != o && t[t[o].fa].rc != o; }
inline void connect(int fa, int o, int d) { t[fa].c[d] = o, t[o].fa = fa; }
inline void pushup(int o) {
    assert(o);
    assert(!t[o].rev);
    for (int i = 0; i < M; ++i)
        t[o].sum[i] = t[t[o].lc].sum[i] + t[t[o].rc].sum[i] + t[o].v[i];
}
inline void pushdown(int o) {
    if (!t[o].rev) return;
    if (t[o].lc) t[t[o].lc].rev ^= 1, std::swap(t[t[o].lc].lc, t[t[o].lc].rc);
    if (t[o].rc) t[t[o].rc].rev ^= 1, std::swap(t[t[o].rc].lc, t[t[o].rc].rc);
    t[o].rev = 0;
}
inline void rotate(int o) {
    assert(!isroot(o));
    int fa = t[o].fa, pa = t[fa].fa, d1 = idtfy(o), d2 = idtfy(fa), b = t[o].c[d1 ^ 1];
    if (!isroot(fa)) t[pa].c[d2] = o; t[o].fa = pa;
    connect(o, fa, d1 ^ 1), connect(fa, b, d1);
    pushup(fa), pushup(o);
    assert(!t[0].lc && !t[0].rc);
}
inline void splay(int o) {
    int x = o, tp = 0;
    st[++tp] = x;
    while (!isroot(x)) st[++tp] = x = t[x].fa;
    while (tp) pushdown(st[tp--]);
    while (!isroot(o)) {
        int fa = t[o].fa;
        if (isroot(fa)) rotate(o);
        else if (idtfy(o) == idtfy(fa)) rotate(fa), rotate(o);
        else rotate(o), rotate(o);
    }
}
inline void access(int o) {
    for (int x = 0; o; o = t[x = o].fa)
        splay(o), t[o].rc = x, pushup(o);
}
inline void mkrt(int o) {
    access(o), splay(o);
    t[o].rev ^= 1, std::swap(t[o].lc, t[o].rc);
}
inline int getrt(int o) {
    access(o), splay(o);
    while (pushdown(o), t[o].lc) o = t[o].lc;
    return splay(o), o;
}
inline void link(int x, int y) {
    mkrt(x);
    if (getrt(y) != x) t[x].fa = y;
    else assert(0);
}
inline void cut(int x, int y) {
    mkrt(x), access(y), splay(y);
    if (t[y].lc == x && !t[x].rc) t[x].fa = t[y].lc = 0, pushup(y);
    else assert(0);
}

inline void work() {
    while (m--) {
        char opt[10];
        scanf("%s", opt);
        if (*opt == 'a') {
            int x, y;
            read(x), read(y);
            ++x, ++y;
            link(x, y);
        } else if (*opt == 'd') {
            int x, y;
            read(x), read(y);
            ++x, ++y;
            cut(x, y);
        } else if (*opt == 'm') {
            double a, b;
            int x, opt;
            read(x), read(opt), scanf("%lf%lf", &a, &b);
            ++x;
            splay(x), t[x].set(opt, a, b), pushup(x);
        } else {
            int x, y;
            double v, vv = 1, ans = 0;
            read(x), read(y), scanf("%lf", &v);
            ++x, ++y;
            if (getrt(x) != getrt(y)) { puts("unreachable"); continue; }
            mkrt(x), access(y), splay(y);
            for (int i = 0; i < M; ++i, vv *= v) ans += vv * t[y].sum[i];
            printf("%.8le\n", ans);
        }
    }
}

inline void init() {
    read(n), read(m);
    fac[0] = 1;
    for (int i = 1; i < M; ++i) fac[i] = fac[i - 1] * i;
    C[0][0] = 1;
    for (int i = 1; i < M; ++i) {
        C[i][0] = 1;
        for (int j = 1; j < M; ++j) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
    int opt;
    double a, b;
    read(opt);
    for (int i = 1; i <= n; ++i) read(opt), scanf("%lf%lf", &a, &b), t[i].set(opt, a, b), pushup(i);
}

int main() {
#ifdef hzhkk
    freopen("hkk.in", "r", stdin);
#endif
    init();
    work();
    fclose(stdin), fclose(stdout);
    return 0;
}
01-02 11:21