【bzoj4756】[Usaco2017 Jan]Promotion Counting

4756: [Usaco2017 Jan]Promotion Counting

Time Limit: 10 Sec   Memory Limit: 128 MB
Submit: 305   Solved: 217
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Description

The cows have once again tried to form a startup company, failing to remember from past experience t

hat cows make terrible managers!The cows, conveniently numbered 1…N1…N (1≤N≤100,000), organize t

he company as a tree, with cow 1 as the president (the root of the tree). Each cow except the presid

ent has a single manager (its "parent" in the tree). Each cow ii has a distinct proficiency rating, 

p(i), which describes how good she is at her job. If cow ii is an ancestor (e.g., a manager of a man

ager 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 seve

ral of her subordinates, in which case the manager should consider promoting some of her subordinate

s. Your task is to help the cows figure out when this is happening. For each cow ii in the company, 

please count the number of subordinates jj where p(j)>p(i).

n只奶牛构成了一个树形的公司，每个奶牛有一个能力值pi，1号奶牛为树根。
问对于每个奶牛来说，它的子树中有几个能力值比它大的。

Input

The first line of input contains N

The next N lines of input contain the proficiency ratings p(1)…p(N) 

for the cows. Each is a distinct integer in the range 1…1,000,000,000

The next N-1 lines describe the manager (parent) for cows 2…N 

Recall that cow 1 has no manager, being the president.

n，表示有几只奶牛 n<=100000
接下来n行为1-n号奶牛的能力值pi
接下来n-1行为2-n号奶牛的经理（树中的父亲）

Output

Please print N lines of output. The ith line of output should tell the number of 

subordinates of cow ii with higher proficiency than cow i.

共n行，每行输出奶牛i的下属中有几个能力值比i大

Sample Input

5
804289384
846930887
681692778
714636916
957747794
1
1
2
3

Sample Output

2
0
1
0
0

HINT

Source

Platinum鸣谢Acty提供译文

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线段树合并

主席树也可
代码：

#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

typedef long long LL;

const int INF = 2147483647;
const int maxn = 100100;
const int maxseg = 20 * maxn;

vector e[maxn];
int n;
int a[maxn],ans[maxn],rt[maxn];
int siz[maxseg],lc[maxseg],rc[maxseg],tot;
int ha[maxn],totha,data[maxn];

inline LL getint()
{
	LL ret = 0,f = 1;
	char c = getchar();
	while (c < '0' || c > '9')
	{
		if (c == '-') f = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9')
		ret = ret * 10 + c - '0',c = getchar();
	return ret * f;
}

inline void maintain(int o)
{
	siz[o] = siz[lc[o]] + siz[rc[o]];
}

inline int merge(int lo,int ro)
{
	if (!lo) return ro;
	if (!ro) return lo;
	int now = lo;
	lc[now] = merge(lc[lo],lc[ro]);
	rc[now] = merge(rc[lo],rc[ro]);
	siz[now] = siz[lo] + siz[ro];
	return now;
}

inline void insert(int &o,int l,int r,int pos)
{
	if (!o) o = ++tot;
	if (l == r) {siz[o]++; return;}
	int mid = l + r >> 1;
	if (pos <= mid) insert(lc[o],l,mid,pos);
	else insert(rc[o],mid + 1,r,pos);
	maintain(o);
}

inline int query(int o,int l,int r,int x)
{
	if (l == r) return 0;
	int mid = l + r >> 1;
	if (x <= mid) return query(lc[o],l,mid,x) + siz[rc[o]];
	else return query(rc[o],mid + 1,r,x);
}

inline void dfs(int u,int fa)
{
	insert(rt[u],1,totha,a[u]);
	for (int i = 0; i < e[u].size(); i++)
	{
		int v = e[u][i];
		if (v == fa) continue;
		dfs(v,u);
		rt[u] = merge(rt[u],rt[v]);
	}
	ans[u] = query(rt[u],1,totha,a[u]);
}

int main()
{
	n = getint();
	for (int i = 1; i <= n; i++)
		data[i] = a[i] = getint();
	for (int i = 2; i <= n; i++)
	{
		int father = getint();
		e[father].push_back(i);
		e[i].push_back(father);
	}
	sort(data + 1,data + n + 1);
	for (int i = 1; i <= n; i++)
		if (data[i] != data[i + 1])
			ha[++totha] = data[i];
	for (int i = 1; i <= n; i++) a[i] = lower_bound(ha + 1,ha + totha + 1,a[i]) - ha;
	dfs(1,0);
	for (int i = 1; i <= n; i++)
		printf("%d\n",ans[i]);
	return 0;
}