Arkady and a Nobody-men CodeForces

大意: 给定有根树, 根节点深度为$1$.

定义$r(a,b)$为$b$子树内深度不超过$a$的节点数$-1$

定义$z_a$为$a$的所有祖先的$r$之和. 对于所有点求出$z$的值.

一个点$y$对$x$的贡献显然为$dep[lca(x,y)]$, 直接计算是$O(n^2logn)$, 考虑优化.

注意到点$y$对$x$子树内所有点贡献都相同, 我们考虑深度相同的点之间的贡献, 就转化为对下面程序的优化.

REP(d,1,n) q[dep[i]].push_back(d);
REP(d,1,n) {
	for (int x:q[d]) for (int y:q[d]) {
		ans[x]+=dep[lca(x,y)];
	}
}

 显然建出虚树然后DP即可, 复杂度为$O(nlogn)$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, rt, fa[N];
vector<int> g[N], q[N], gg[N];
int sz[N], dep[N], son[N];
int L[N], R[N], top[N];
ll ans[N], c[N];

void dfs(int x, int d) {
	L[x]=++*L,dep[x]=d,sz[x]=1;
	for (int y:g[x]) {
		dfs(y,d+1),sz[x]+=sz[y];
		if (sz[y]>sz[son[x]]) son[x]=y;
	}
	R[x]=*L;
}
void dfs2(int x, int tf) {
	top[x]=tf;
	if (son[x]) dfs2(son[x],tf);
	for (int y:g[x]) if (!top[y]) dfs2(y,y);
}
int lca(int x, int y) {
	while (top[x]!=top[y]) {
		if (dep[top[x]]<dep[top[y]]) swap(x,y);
		x = fa[top[x]];
	}
	return dep[x]<dep[y]?x:y;
}
void dfs(int x) {
	for (int y:g[x]) ans[y]+=ans[x],dfs(y);
}
int s[N], cnt;
void DP1(int x) {
    sz[x]=gg[x].empty();
    for (int y:gg[x]) DP1(y),sz[x]+=sz[y];
}
void DP2(int x) {
    for (int y:gg[x]) c[y]=(ll)dep[x]*(sz[x]-sz[y])+c[x],DP2(y);
}
void solve(vector<int> a) {
	int sz = a.size();
	sort(a.begin(),a.end(),[](int a,int b){return L[a]<L[b];});
	REP(i,1,sz-1) a.pb(lca(a[i],a[i-1]));
	sort(a.begin(),a.end(),[](int a,int b){return L[a]<L[b];});
	a.erase(unique(a.begin(),a.end()),a.end());
	s[cnt=1]=a[0], sz = a.size();
	REP(i,1,sz-1) {
		while (cnt>=1) {
			if (L[s[cnt]]<=L[a[i]]&&L[a[i]]<=R[s[cnt]]) {
				gg[s[cnt]].pb(a[i]);
				break;
			}
			--cnt;
		}
		s[++cnt]=a[i];
	}
	DP1(s[1]),c[s[1]]=0,DP2(s[1]);
	for (int x:a) gg[x].clear();
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d",fa+i),g[fa[i]].pb(i);
	rt = min_element(fa+1,fa+1+n)-fa;
	dfs(rt,1),dfs2(rt,rt);
	REP(i,1,n) q[dep[i]].pb(i);
	REP(d,1,n) if (q[d].size()) { 
		solve(q[d]);
		for (int x:q[d]) ans[x]=c[x]+dep[x];
	}
	dfs(rt);
	REP(i,1,n) printf("%lld ", ans[i]-dep[i]);hr;
}