P3605 [USACO17JAN]Promotion Counting晋升者计数

题目描述

The cows have once again tried to form a startup company, failing to remember from past experience that cows make terrible managers!

The cows, conveniently numbered $\ldots N1…N ( 1 \leq N \leq 100,0001≤N≤100,000 ), organize the company as a tree, with cow 1 as the president (the root of the tree). Each cow except the president has a single manager (its "parent" in the tree). Each cow ii has a distinct proficiency rating, p(i)p(i) , which describes how good she is at her job. If cow ii is an ancestor (e.g., a manager of a manager of a manager) of cow jj , then we say jj is a subordinate of ii .$

Unfortunately, the cows find that it is often the case that a manager has less proficiency than several of her subordinates, in which case the manager should consider promoting some of her subordinates. Your task is to help the cows figure out when this is happening. For each cow

奶牛们又一次试图创建一家创业公司，还是没有从过去的经验中吸取教训--牛是可怕的管理者！

为了方便，把奶牛从

不幸地是，奶牛们发现经常发生一个上司比她的一些下属能力低的情况，在这种情况下，上司应当考虑晋升她的一些下属。你的任务是帮助奶牛弄清楚这是什么时候发生的。简而言之，对于公司的中的每一头奶牛

输入输出格式

输入格式：

The first line of input contains

The next $\ldots p(N)p(1)…p(N) for the cows. Each is a distinct integer in the range 1 \ldots 1,000,000,0001…1,000,000,000 .$

The next $\ldots N2…N . Recall that cow 1 has no manager, being the president.$

输入的第一行包括一个整数

接下来的 $\cdots p(N)p(1)⋯p(N) . 保证所有数互不相同，在区间 1 \cdots 10^91⋯109 之间。$

接下来的 $\cdots N2⋯N 的上司（双亲节点）的编号。再次提醒，1 号奶牛作为总裁，没有上司。$

输出格式：

Please print

输出包括

输入输出样例

输入样例#1：复制

输出样例#1：复制

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总结：线段树合并裸题
dfs时，将每一颗子树对应的权值线段树合并到当前节点，然后查询
并查集维护连通性

#include <bits/stdc++.h>
using namespace std;

const int maxn = 4e5 + 7;
int n, m, a[maxn], b[maxn], tot, cnt = 1;
int head[maxn], sum;
struct Node {
	int v, nxt;
} G[maxn];
struct V {
	int v, ls, rs;
} t[maxn];

int f[maxn], rt[maxn], ans[maxn];
void ins(int u, int v) {
	G[cnt] = (Node) {v, head[u]}; head[u] = cnt++;
}
int find(int x) {
	return f[x] == x ?x :f[x] = find(f[x]);
} 
void Modify(int &o, int l, int r, int w) {
	if(!o) o = ++tot; t[o].v++;
	if(l == r) return;
	int mid = (l + r) >> 1;
	if(w <= mid) Modify(t[o].ls, l, mid, w);
	else Modify(t[o].rs, mid + 1, r, w);
}
void Merge(int &r1, int r2) {
	if(!r1) {r1 = r2; return;}
	if(!r2) return;
	t[r1].v += t[r2].v;
	Merge(t[r1].ls, t[r2].ls); Merge(t[r1].rs, t[r2].rs);
}
int Query(int o, int l, int r, int ql, int qr) {
	if(ql <= l && r <= qr) return t[o].v;
	int mid = (l + r) >> 1, res = 0;
	if(ql <= mid) res += Query(t[o].ls, l, mid, ql, qr);
	if(qr > mid) res += Query(t[o].rs, mid + 1, r, ql, qr);
	return res;
} 
void dfs(int x, int fa) {
	for (int i = head[x]; i; i = G[i].nxt) {
		int v = G[i].v; if(v == fa) continue;
		dfs(v, x); int r1 = find(x), r2 = find(v);
		if(r1 != r2) {
			f[r2] = r1; Merge(rt[r1], rt[r2]);
		}
	}
	ans[x] = Query(rt[find(x)], 1, sum, a[x] + 1, sum);
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]); b[i] = a[i];
	}
	for (int i = 2; i <= n; ++i) {
		int x; scanf("%d", &x); ins(x, i); ins(i, x);
	}
	sort(b + 1, b + n + 1); sum = unique(b + 1, b + n + 1) - b - 1;
	for (int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + sum + 1, a[i]) - b;
	for (int i = 1; i <= n; ++i) {
		Modify(rt[f[i] = i], 1, sum, a[i]);
	} dfs(1, 0);
	for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
	return 0;
}