题目链接
题解
观察题目的式子似乎没有什么意义,我们考虑计算出每一种权值的概率
先离散化一下权值
显然可以设一个\(dp\),设\(f[i][j]\)表示\(i\)节点权值为\(j\)的概率
如果\(i\)是叶节点显然
如果\(i\)只有一个儿子直接继承即可
如果\(i\)有两个儿子,对于儿子\(x\),设另一个儿子为\(y\)
则有
\[f[i][j] += f[x][j](1 - p_i)\sum\limits_{k > j}f[r][k] + f[x][j]p_i\sum\limits_{k < j}f[r][k]
\]
\]
直接转移是\(O(n^2)\)的,发现每个节点都有\(O(n)\)个位置需要转移
考虑优化,可以考虑线段树合并
对于一个子树中的权值\(x\),我们记另一棵子树比它大的概率为\(maxa\),
则\(x\)的概率要乘上\(maxa(1 - p_i) + (1 - maxa)p_i = maxa + p_i - 2p_imaxa\)
所以我们在线段树合并过程中,优先合并右子树,并更新两棵子树的\(maxa\)与\(maxb\),就可以在合并过程中转移了
复杂度\(O(nlogn)\)
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 300005,maxm = 8000005,INF = 1000000000,P = 998244353;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int n,Ls[maxn],Rs[maxn],b[maxn],N,v10000;
int rt[maxn],sum[maxm],ls[maxm],rs[maxm],tag[maxm],cnt;
int p[maxn],maxa,maxb;
inline int qpow(int a,int b){
int re = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) re = 1ll * re * a % P;
return re;
}
inline void pd(int u){
if (tag[u] > 1){
sum[ls[u]] = 1ll * sum[ls[u]] * tag[u] % P;
sum[rs[u]] = 1ll * sum[rs[u]] * tag[u] % P;
tag[ls[u]] = 1ll * tag[ls[u]] * tag[u] % P;
tag[rs[u]] = 1ll * tag[rs[u]] * tag[u] % P;
tag[u] = 1;
}
}
void modify(int& u,int l,int r,int pos){
u = ++cnt; sum[u] = tag[u] = 1;
if (l == r) return;
int mid = l + r >> 1;
if (mid >= pos) modify(ls[u],l,mid,pos);
else modify(rs[u],mid + 1,r,pos);
}
int merge(int u,int v,int p){
if (!u && !v) return 0;
if (!u){
maxb = (maxb + sum[v]) % P;
int tmp;
tmp = (((maxa + p) % P - 2ll * p * maxa % P) % P + P) % P;
sum[v] = 1ll * sum[v] * tmp % P;
tag[v] = 1ll * tag[v] * tmp % P;
return v;
}
if (!v){
maxa = (maxa + sum[u]) % P;
int tmp;
tmp = (((maxb + p) % P - 2ll * p * maxb % P) % P + P) % P;
sum[u] = 1ll * sum[u] * tmp % P;
tag[u] = 1ll * tag[u] * tmp % P;
return u;
}
pd(u); pd(v);
int t = ++cnt; tag[t] = 1;
rs[t] = merge(rs[u],rs[v],p);
ls[t] = merge(ls[u],ls[v],p);
sum[t] = (sum[ls[t]] + sum[rs[t]]) % P;
return t;
}
void dfs(int u){
if (!Ls[u]) modify(rt[u],1,N,p[u]);
else if (!Rs[u]) dfs(Ls[u]),rt[u] = rt[Ls[u]];
else {
dfs(Ls[u]); dfs(Rs[u]);
maxa = maxb = 0;
rt[u] = merge(rt[Ls[u]],rt[Rs[u]],p[u]);
}
}
int ans;
void cal(int u,int l,int r){
if (l == r) {ans = (ans + 1ll * l * b[l] % P * sum[u] % P * sum[u] % P) % P;return;}
pd(u);
int mid = l + r >> 1;
cal(ls[u],l,mid);
cal(rs[u],mid + 1,r);
}
int main(){
n = read(); read(); int x; v10000 = qpow(10000,P - 2);
for (int i = 2; i <= n; i++){
x = read();
if (!Ls[x]) Ls[x] = i;
else Rs[x] = i;
}
for (int i = 1; i <= n; i++){
p[i] = read();
if (!Ls[i]) b[++N] = p[i];
else p[i] = 1ll * p[i] * v10000 % P;
}
sort(b + 1,b + 1 + N);
for (int i = 1; i <= n; i++)
if (!Ls[i]) p[i] = lower_bound(b + 1,b + 1 + N,p[i]) - b;
dfs(1);
cal(rt[1],1,N);
printf("%d\n",ans);
return 0;
}