(mathcal{Description})
Link.
给定一棵包含 (n) 个点,有点权和边权的树。设当前位置 (s)(初始时 (s=1)),每次在 (n) 个结点内随机选择目标结点 (t),付出「(s) 到 (t) 的简单路径上的边权之和」( imes)「(t) 的点权」的代价,标记(可以重复标记)点 (t) 并把 (s) 置为 (t)。求每个点至少被标记一次时(其中 (1) 号结点一开始就被标记)代价之和的期望。答案对 (998244353) 取模。
(nle10^6)。
(mathcal{Solution})
首先,有期望意义下的 Min-Max 容斥的公式:
对于本题,(S={2,3,cdots,n}),要求的答案等价于标记最后一个未标记点时代价的期望,那么枚举上式中的 (T),并设 (|T|=m),我们只需要对于每个 (T),求出从结点 (1) 出发,标记 (T) 集合内任意一个点的期望代价就行。
考虑一个朴素的 DP:令 (f_T(u)) 表示现在在 (u) 点((u) 点已标记)时,标记 (T) 内任意一点的期望代价。显然:
其中 (operatorname{dist}(u,v)) 即表示题意中把 (u) 置为 (v) 的代价。到此,你就可以获得 (10) 分的好成绩啦!
接下来,取出一个 (u otin T) 的 (f_T(u)) 来研究:
令 (s=sum_{v otin T}f(v)),(w(u)=sum_{v=1}^noperatorname{dist}(u,v)),(C=S-T={c_1,c_2,cdots,c_{n-m}})(其中恒有 (c_1=1))列出共 (|C|) 个等式:
左右分别相加得到:
于是乎,要求的 (f_T(1)) 就有:
此后,把 (f_T(1)) 带入答案的式子里:
令 (g(m)=sum_{Tsubseteq Sland|T|=m}sum_{v otin T}w(v)),单独考虑结点 (1),它必然不属于 (T);再考虑其他结点的贡献次数,可以得出:
最后,只需要求出 (sum_{u=1}^nw(u))。分别考虑每条边 ((u,v,b)in E) 的贡献。将这条边删去,记此时 (u) 所在联通块的结点个数为 (p_u),结点点权之和为 (q_u),(v) 同理。则:
综上,求出这一系列式子,问题就以 (mathcal O(n)) 的复杂度解决啦!
(mathcal{Code})
/* Clearink */
#include <cstdio>
inline char fgc () {
static char buf[1 << 17], *p = buf, *q = buf;
return p == q && ( q = buf + fread ( p = buf, 1, 1 << 17, stdin ), p == q ) ? EOF : *p++;
}
inline int rint () {
int x = 0; char s = fgc ();
for ( ; s < '0' || '9' < s; s = fgc () );
for ( ; '0' <= s && s <= '9'; s = fgc () ) x = x * 10 + ( s ^ '0' );
return x;
}
const int MAXN = 1e6, MOD = 998244353;
int n, ecnt, head[MAXN + 5], val[MAXN + 5], fac[MAXN + 5], ifac[MAXN + 5];
int fa[MAXN + 5], siz[MAXN + 5], sum[MAXN + 5], dist[MAXN + 5]; // dist[i]==dist(1,i).
int g[MAXN + 5];
inline int mul ( const long long a, const int b ) { return a * b % MOD; }
inline int sub ( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }
inline int add ( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline void addeq ( int& a, const int b ) { ( a += b ) >= MOD && ( a -= MOD, 0 ); }
struct Edge { int to, cst, nxt; } graph[MAXN * 2 + 5];
inline void link ( const int s, const int t, const int c ) {
graph[++ecnt] = { t, c, head[s] };
head[s] = ecnt;
}
inline int qkpow ( int a, int b ) {
int ret = 1;
for ( ; b; a = mul ( a, a ), b >>= 1 ) ret = mul ( ret, b & 1 ? a : 1 );
return ret;
}
inline void init () {
fac[0] = 1;
for ( int i = 1; i <= n; ++i ) fac[i] = mul ( i, fac[i - 1] );
ifac[n] = qkpow ( fac[n], MOD - 2 );
for ( int i = n - 1; ~i; -- i ) ifac[i] = mul ( i + 1ll, ifac[i + 1] );
}
inline int inv ( const int x ) { return mul ( fac[x - 1], ifac[x] ); }
inline int comb ( const int n, const int m ) {
return n < m ? 0 : mul ( fac[n], mul ( ifac[m], ifac[n - m] ) );
}
inline void dfs ( const int u ) {
siz[u] = 1, sum[u] = val[u];
for ( int i = head[u], v; i; i = graph[i].nxt ) {
if ( ( v = graph[i].to ) ^ fa[u] ) {
dist[v] = add ( dist[fa[v] = u], graph[i].cst );
dfs ( v ), siz[u] += siz[v], addeq ( sum[u], sum[v] );
}
}
}
int main () {
n = rint ();
int vs = 0;
for ( int i = 1; i <= n; ++i ) vs = add ( vs, val[i] = rint () );
for ( int i = 1, u, v, w; i < n; ++i ) {
u = rint (), v = rint (), w = rint ();
link ( u, v, w ), link ( v, u, w );
}
init (), dfs ( 1 );
int S = 0;
for ( int u = 1; u <= n; ++u ) {
for ( int i = head[u], v; i; i = graph[i].nxt ) {
if ( ( v = graph[i].to ) ^ fa[u] ) {
addeq ( S, mul ( graph[i].cst,
add ( mul ( siz[v], sub ( vs, sum[v] ) ),
mul ( n - siz[v], sum[v] ) ) ) );
}
}
}
for ( int i = 2; i <= n; ++i ) addeq ( g[n - 1], mul ( val[i], dist[i] ) );
for ( int i = n - 2; i; --i ) {
g[i] = add ( mul ( comb ( n - 1, i ), g[n - 1] ),
mul ( comb ( n - 2, i ), sub ( S, g[n - 1] ) ) );
}
int ans = 0;
for ( int i = 1; i < n; ++i ) {
ans = ( i & 1 ? add : sub )( ans,
add ( mul ( comb ( n - 1, i ), g[n - 1] ), mul ( inv ( i ), g[i] ) ) );
}
printf ( "%d
", mul ( ans, inv ( n ) ) );
return 0;
}