P3605 [USACO17JAN]Promotion Counting P

Jennie

和常规的求逆序对差不多

在从根节点往下走的时候，我们必须要避免不在他子树内的点的影响

那就先减去他们呗。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
template<class T>
void read(T &now){
	now=0;
	char c=getchar();
	int f=1;
	while((!isdigit(c))){
		if(c=='-') f=-1;
	//	cout<<isdigit(c)<<" "<<c<<" ";
		c=getchar();
	}
	while(isdigit(c)){
		now=(now<<1)+(now<<3)+c-'0';
		c=getchar();
	}
	now*=f;
}
int n;
int b[1000005];
int p[1000005];
int head[1000005];
int ans[1000005];
struct e{
	int to;
	int ne;
}ed[1000005];
int x;
int pp;
void add(int f,int to){
	ed[++pp].to=to;
	ed[pp].ne=head[f];
	head[f]=pp;
}
int ma;
int lowbit(int x){
	return x&-x;
}
int tr[1000005];
void addd(int x,int k){
	for(int i=x;i<=n;i+=lowbit(i)){
		tr[i]+=k;
	}
}
int qu(int x){
	int ans=0;
	while(x){
		ans+=tr[x];
		x-=lowbit(x);
	}
	return ans;
}
void dfs(int x){
	ans[x]-=(qu(n)-qu(p[x]));
	for(int i=head[x];i;i=ed[i].ne){
		dfs(ed[i].to);
	}
	ans[x]+=(qu(n)-qu(p[x]));
	addd(p[x],1);
	//cout<<qu(n)<<endl;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		read(b[i]);
		p[i]=b[i];
	}
	sort(b+1,b+n+1);
	for(int i=1;i<=n;++i){
		p[i]=lower_bound(b+1,b+n+1,p[i])-b;
	}
	for(int i=2;i<=n;++i){
		read(x);
		add(x,i);
	}
	dfs(1);
	for(int i=1;i<=n;++i){
		printf("%d\n",ans[i]);
	}
	return 0;
}