[PKUWC2018]Minimax [dp,线段树合并]

好妙的一个题…

我们设 (f_{i,j}) 为 (i) 节点出现 (j) 的概率

设 (l = ch[i][0] , r = ch[i][1])
即左儿子右儿子

设 (m) 为叶子结点的个数

显然，(i) 出现 (j) 的概率为

[f_{i,j} = f_{l,j} * (p_i sum_{k=1}^{j-1}f_{r,k} + (1-p_i)sum_{k=j+1}^{m}f_{r,k}) + f_{r,j} * (p_i sum_{k=1}^{j-1}f_{l,k} + (1-p_i)sum_{k=j+1}^{m}f_{l,k}) ]

不难发现，这个柿子有关前缀和和后缀和，可以用线段树合并的操作来进行转移，从下到上转移，求出根节点的概率就好了…

#include <cstdio>
#include <algorithm>

int read() {
  int x = 0;
  char c = 0;
  while (c < 48) c = getchar();
  while (c > 47) x = (x << 1) + (x << 3) + (c & 15), c = getchar();
  return x;
}

const int mod = 998244353;
int qpow(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) ans = 1ll * ans * x % mod;
  return ans;
}

int n;
const int maxn = 3e5 + 10;
int ch[maxn][2], fa[maxn], cnt[maxn], val[maxn], tmp[maxn], qwq = 0, s[maxn];
int rt[maxn], ls[maxn << 5], rs[maxn << 5], sum[maxn << 5], mul[maxn << 5];
int ans = 0, tot = 0;

void pushup(int rt) { sum[rt] = (sum[ls[rt]] +sum[rs[rt]]) % mod; }
void pushmul(int rt, int v) {
  if (!rt) return;
  sum[rt] = 1ll * sum[rt] * v % mod;
  mul[rt] = 1ll * mul[rt] * v % mod;
}

void pushd(int rt) {
  if (mul[rt] == 1) return;
  if (ls[rt]) pushmul(ls[rt], mul[rt]);
  if (rs[rt]) pushmul(rs[rt], mul[rt]);
  mul[rt] = 1;
}

int newnode() {
	int x = ++ tot; 
	ls[x] = rs[x] = sum[x] = 0, mul[x] = 1 ;
	return x ;
}
void upd(int& p, int l, int r, int x, int v) {
  if (!p) p = newnode() ;
  if (l == r) {
    sum[p] = v;
    return;
  }
  pushd(p);
  int mid = l + r >> 1;
  (x <= mid) ? upd(ls[p], l, mid, x, v) : upd(rs[p], mid + 1, r, x, v);
  pushup(p);
}

int merge(int x, int y, int l, int r, int xmul, int ymul, int v) {
  if (!x && !y) return 0;
  if (!x) {
    pushmul(y, ymul);
    return y;
  }
  if (!y) {
    pushmul(x, xmul);
    return x;
  }
  pushd(x), pushd(y);
  int mid = l + r >> 1;
  int lsx = sum[ls[x]], lsy = sum[ls[y]], rsx = sum[rs[x]], rsy = sum[rs[y]];
  ls[x] = merge(ls[x], ls[y], l, mid, (xmul + 1ll * rsy % mod * (1 - v + mod)) % mod,
                (ymul + 1ll * rsx % mod * (1 - v + mod)) % mod, v);
  rs[x] = merge(rs[x], rs[y], mid + 1, r, (xmul + 1ll * lsy % mod * v) % mod,
                (ymul + 1ll * lsx % mod * v) % mod, v);
  pushup(x);
  return x;
}

void out(int x, int l, int r) {
  if (!x) return;
  if (l == r) {
    s[l] = sum[x];
    return;
  }
  int mid = l + r >> 1;
  pushd(x);
  out(ls[x], l, mid);
  out(rs[x], mid + 1, r);
}

void dfs(int u) {
  if (!cnt[u]) upd(rt[u], 1, qwq, val[u], 1);
  if (cnt[u] == 1) dfs(ch[u][0]), rt[u] = rt[ch[u][0]] ;
  if (cnt[u] == 2) dfs(ch[u][0]), dfs(ch[u][1]), rt[u] = merge(rt[ch[u][0]], rt[ch[u][1]] ,1 , qwq , 0 , 0 , val[u]);
}

int main() {
  n = read();
  for (int i = 1; i <= n; i++) fa[i] = read();
  for (int i = 1; i <= n; i++)
    if (fa[i]) ch[fa[i]][cnt[fa[i]]++] = i;
  for (int i = 1; i <= n; i++) val[i] = read();
  for (int i = 1; i <= n; i++) {
    if (cnt[i]) {
      val[i] = 1ll * val[i] * qpow(10000, mod - 2) % mod;
    } else {
      tmp[++qwq] = val[i];
    }
  }
  std ::sort(tmp + 1, tmp + qwq + 1);
  for (int i = 1; i <= n; i++)
    if (!cnt[i]) val[i] = std ::lower_bound(tmp + 1, tmp + qwq + 1, val[i]) - tmp;
  dfs(1);
  out(rt[1], 1, qwq);
  for (int i = 1; i <= qwq; i++) ans = (ans + 1ll * i * tmp[i] % mod * s[i] % mod * s[i]) % mod;
  printf("%d
", ans);
  return 0;
}