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【题解】HackerRank - Fibonacci Numbers Tree

题目

https://www.hackerrank.com/challenges/fibonacci-numbers-tree/

题意

给一棵树,要求支持两种操作:

  • U x k 给 x 所在的子树中的每一个结点 y 加上一个值 $Fib(k+dist(x,y))$。
  • Q x y询问 x 到 y 的路径上所有结点的值的和。

题解

  • 如果子树上所有结点都加上一个定值的话,那么就能做了。只需要 树链剖分+线段树 就好了。

  • 利用 Fibonacci 数列的一个性质 $F_{a+b}=F_{a-1} \cdot F_b+F_a \cdot F_{b+1}$(对于负数项同样成立),将 $F(k+dist(x,y))$ 拆分为:

    $$\begin{eqnarray}F(k-dep(x)+dep(y)) &=& F(k-dep(x)-1) \cdot F(dep(y))+F(k-dep(x)) \cdot F(dep(y)+1) \ &=& C_1 \cdot F(dep(y))+C_2 \cdot F(dep(y)+1)\end{eqnarray}$$

  • 然后。。。然后就能做了嘛!

代码(套了一堆板子,有点长)

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#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args << " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... Args>
void err(T<t> a, Args... args) { for (auto x: a) cout << x << ' '; err(args...); }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int maxn = 1E5 + 100;
const LL MOD = 1E9 + 7;
vector<int> G[maxn];
int dep[maxn], idx[maxn], sz[maxn], son[maxn], top[maxn], dep_idx[maxn], fa[maxn], clk, out[maxn];
int n, Q;
LL fib[maxn] = {0, 1};

void add(LL &a, LL b) { (a += b) %= MOD; }

namespace BM {
inline void up(LL& a, LL b) { (a += b) %= MOD; }
typedef vector<LL> V;
V mul(const V& a, const V& b, const V& m, int k) {
V r; r.resize(2 * k - 1);
FOR (i, 0, k)
FOR (j, 0, k)
up(r[i + j], a[i] * b[j]);
FORD (i, k - 2, - 1) {
FOR (j, 0, k)
up(r[i + j], r[i + k] * m[j]);
r.pop_back();
}
return r;
}

V pow(LL n, const V& m) {
int k = (int)m.size() - 1; assert(m[k] == -1 || m[k] == MOD - 1);
V r(k), x(k); r[0] = x[1] = 1;
for (; n; n >>= 1, x = mul(x, x, m, k))
if (n & 1) r = mul(x, r, m, k);
return r;
}

LL go(LL n) {
static V a = {1, 1, -1};
static V x = {1, 1};
static int k = 2;
if (n <= k) return x[n - 1];
V r = pow(n - 1, a);
LL ans = 0;
FOR (i, 0, k)
up(ans, r[i] * x[i]);
return ans;
}

LL fib(LL n) {
if (n == 0) return 0;
if (n > 0) return go(n);
return (n & 1) ? go(-n) : MOD - go(-n);
}
}

namespace TREE {
struct Q {
LL v0, v1;
Q(LL d0 = 0, LL d1 = 0): v0(d0), v1(d1) {}
void operator += (Q& q) { add(v0, q.v0); add(v1, q.v1); }
};
struct P {
LL sum;
P(LL sum = 0): sum(sum) {}
void up(Q& q, LL d0, LL d1) { add(sum, d0 * q.v0 + d1 * q.v1); }
};
template<typename T>
P operator & (T&& a, T&& b) {
return P((a.sum + b.sum) % MOD);
}
P p[maxn << 2];
Q q[maxn << 2];
LL d0[maxn << 2], d1[maxn << 2];
#define ls o * 2, l, (l + r) / 2
#define rs o * 2 + 1, (l + r) / 2 + 1, r
void build(int o, int l, int r) {
if (l == r) {
d0[o] = fib[dep_idx[l]];
d1[o] = fib[dep_idx[l] + 1];
return;
}
build(ls); build(rs);
add(d0[o], d0[o * 2] + d0[o * 2 + 1]);
add(d1[o], d1[o * 2] + d1[o * 2 + 1]);
};
void maintain(int o, int l, int r) {
if (l == r) p[o] = P();
else p[o] = p[o * 2] & p[o * 2 + 1];
p[o].up(q[o], d0[o], d1[o]);
}
void pushdown(int o, int l, int r) {
q[o * 2] += q[o]; q[o * 2 + 1] += q[o];
q[o] = Q();
maintain(ls); maintain(rs);
}
P query(int ql, int qr, int o = 1, int l = 1, int r = n) {
if (ql > r || l > qr) return P();
if (ql <= l && r <= qr) return p[o];
pushdown(o, l, r);
return query(ql, qr, ls) & query(ql, qr, rs);
}
void update(int ql, int qr, Q v, int o = 1, int l = 1, int r = n) {
if (ql > r || l > qr) return;
if (ql <= l && r <= qr) q[o] += v;
else {
pushdown(o, l, r);
update(ql, qr, v, ls); update(ql, qr, v, rs);
}
maintain(o, l, r);
}
}


void predfs(int u, int d) {
dep[u] = d;
sz[u] = 1;
int& maxs = son[u] = -1;
for (int v: G[u])
if (v != fa[u]) {
fa[v] = u;
predfs(v, d + 1);
sz[u] += sz[v];
if (maxs == -1 || sz[v] > sz[maxs])
maxs = v;
}
}

void dfs(int u, int tp) {
top[u] = tp;
idx[u] = ++clk;
dep_idx[idx[u]] = dep[u];
if (son[u] != -1) dfs(son[u], tp);
for (int v: G[u])
if (v != son[u] && v != fa[u])
dfs(v, v);
out[u] = clk;
}

LL query(int u, int v) {
int uu = top[u], vv = top[v];
LL ret = 0;
while (uu != vv) {
if (dep[uu] < dep[vv]) { swap(uu, vv); swap(u, v); }
add(ret, TREE::query(idx[uu], idx[u]).sum);
u = fa[uu];
uu = top[u];
}
if (dep[u] < dep[v]) swap(u, v);
add(ret, TREE::query(idx[v], idx[u]).sum);
return ret;
}


int main() {
#ifdef zerol
freopen("in", "r", stdin);
// freopen("out", "w", stdout);
#endif
FOR (i, 2, maxn) fib[i] = (fib[i - 1] + fib[i - 2]) % MOD;
cin >> n >> Q;
FOR (u, 2, n + 1) {
int v; scanf("%d", &v);
G[v].push_back(u); G[u].push_back(v);
}
predfs(1, 1); dfs(1, 1); TREE::build(1, 1, n);
char s[10];
while (Q--) {
scanf("%s", s);
if (s[0] == 'U') {
int u; LL k; scanf("%d%lld", &u, &k);
TREE::update(idx[u], out[u], {BM::fib(k - dep[u] - 1), BM::fib(k - dep[u])});
} else if (s[0] == 'Q') {
int u, v; scanf("%d%d", &u, &v);
printf("%lld\n", query(u, v));
}
}
}