「PKUWC2018」Minimax

题面

题解

强势安利一波巨佬的$blog$

线段树合并吼题啊

合并的时候要记一下$A$点权值小于$l$的概率和$A$点权值大于$r$的概率，对$B$点同样做

时空复杂度$\text O(nlogw)$，$w$为其中权值的最大值

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int Mod(998244353), Inv(796898467), maxn(300010), LIM((1 << 30) - 1);
inline int Add(int a, int b) { return (a + b) % Mod; }
int ans, cnt, son[2][maxn * 20], tag[maxn * 20], P[maxn * 20], cur, root[maxn];

inline int newNode()
{
	tag[++cur] = 1; P[cur] = 0;
	son[0][cur] = son[1][cur] = 0;
	return cur;
}

inline void pushup(int x) { P[x] = Add(P[son[0][x]], P[son[1][x]]); }
inline void put_tag(int x, int p)
{
	if(!x) return;
	tag[x] = 1ll * tag[x] * p % Mod;
	P[x] = 1ll * P[x] * p % Mod;
}

inline void pushdown(int x)
{
	put_tag(son[0][x], tag[x]);
	put_tag(son[1][x], tag[x]);
	tag[x] = 1;
}

void calc(int x, int l = 0, int r = LIM)
{
	if(!x) return;
	if(l == r)
	{
		++cnt; ans = Add(ans, 1ll * cnt * P[x] % Mod * P[x] % Mod * l % Mod);
		return;
	}
	pushdown(x);
	int mid = (l + r) >> 1;
	calc(son[0][x], l, mid);
	calc(son[1][x], mid + 1, r);
}

void insert(int &x, int id, int l = 0, int r = LIM)
{
	if(!x) x = newNode();
	tag[x] = P[x] = 1;
	if(l == r) return;
	int mid = (l + r) >> 1;
	if(id <= mid) insert(son[0][x], id, l, mid);
	else insert(son[1][x], id, mid + 1, r);
}

int S[2][maxn], tot[maxn], W[maxn], n;
int merge(int a, int b, int pa, int pb, int pmax)
{
	if(!a && !b) return 0;
	if(!a) return put_tag(b, pa), b;
	if(!b) return put_tag(a, pb), a;
	pushdown(a), pushdown(b);
	int pal = Add(pa, 1ll * P[son[1][a]] * Add(Mod - pmax, 1) % Mod),
		pbl = Add(pb, 1ll * P[son[1][b]] * Add(Mod - pmax, 1) % Mod),
		par = Add(pa, 1ll * P[son[0][a]] * pmax % Mod),
		pbr = Add(pb, 1ll * P[son[0][b]] * pmax % Mod);
	son[0][a] = merge(son[0][a], son[0][b], pal, pbl, pmax);
	son[1][a] = merge(son[1][a], son[1][b], par, pbr, pmax);
	return pushup(a), a;
}

void dfs(int x)
{
	if(!tot[x]) return (void) (insert(root[x], W[x]));
	W[x] = 1ll * W[x] * Inv % Mod;
	for(RG int i = 0; i < tot[x]; i++) dfs(S[i][x]);
	if(tot[x] == 1) root[x] = root[S[0][x]];
	else root[x] = merge(root[S[0][x]], root[S[1][x]], 0, 0, W[x]);
}

int main()
{
#ifndef ONLINE_JUDGE
	file(cpp);
#endif
	n = read();
	for(RG int i = 1, fa; i <= n; i++)
		fa = read(), S[tot[fa]++][fa] = i;
	for(RG int i = 1; i <= n; i++) W[i] = read();
	dfs(1); calc(root[1]); printf("%d\n", ans);
	return 0;
}