#树上启发式合并，trie#JZOJ 5363 生命之树

---

分析

考虑按位处理，
如果熟悉dsu的话可以发现这道题能够用dsu做，
再用两个trie分别维护该位为0或1的字符串，
重儿子可以按照子树字符串的总长计算

---

代码

#include <cstdio>
#include <cctype>
#include <cstring>
#define rr register
using namespace std;
const int N=100011,M=500011; typedef long lll; char s[M];
struct node{int y,next;}e[N<<1]; lll now,ans[N],siz[N];
int as[N],a[N],S,et=1,n,r[N],root,dep[N],fat[N],big[N];
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
inline void print(lll ans){
	if (ans>9) print(ans/10);
	putchar(ans%10+48);
}
struct Trie{
	int tot,trie[M][26],cnt[M];
	inline void Clear(){memset(trie[1],0,sizeof(trie[1])),cnt[tot=1]=0;}
	inline void Insert(int L,int R){
		rr int p=1; ++cnt[p];
		for (rr int i=L;i<R;++i){
			if (!trie[p][s[i]-97]){
			    trie[p][s[i]-97]=++tot,cnt[tot]=0,
				memset(trie[tot],0,sizeof(trie[tot]));
			}
			p=trie[p][s[i]-97],++cnt[p];
		}
	}
	inline signed query(int L,int R){
		rr int p=1,sum=0;
		for (rr int i=L;i<R;++i){
			p=trie[p][s[i]-97];
			if (!p) return sum;
			sum+=cnt[p];
		}
		return sum;
	}
}trie0,trie1;
inline void dfs1(int x,int fa){
	dep[x]=dep[fa]+1,fat[x]=fa,siz[x]=r[x]-r[x-1];
	for (rr int i=as[x],SIZ=-1;i;i=e[i].next)
	if (e[i].y!=fa){
		dfs1(e[i].y,x);
		siz[x]+=siz[e[i].y];
		if (siz[e[i].y]>SIZ) big[x]=e[i].y,SIZ=siz[e[i].y];
	}
}
inline void updqry(int x){
	for (rr int i=as[x];i;i=e[i].next)
	    if (e[i].y!=fat[x]&&e[i].y!=root) updqry(e[i].y);
	if (a[x]&S) now+=trie0.query(r[x-1],r[x]),trie1.Insert(r[x-1],r[x]);
	    else now+=trie1.query(r[x-1],r[x]),trie0.Insert(r[x-1],r[x]);
}
inline void dfs2(int x,int opt){
	for (rr int i=as[x];i;i=e[i].next)
	    if (e[i].y!=fat[x]&&e[i].y!=big[x]) dfs2(e[i].y,0);
	if (big[x]) dfs2(big[x],1),root=big[x];
	updqry(x),ans[x]+=now*S,root=0;
	if (!opt) trie0.Clear(),trie1.Clear(),now=0;
}
signed main(){
	freopen("tree.in","r",stdin);
	freopen("tree.out","w",stdout);
    n=iut(),trie0.tot=trie1.tot=r[0]=1;
    for (rr int i=1;i<=n;++i) a[i]=iut();
    for (rr int i=1;i<=n;++i){
    	r[i]=r[i-1]; rr char c=getchar();
    	while (!isalpha(c)) c=getchar();
    	while (isalpha(c)) s[r[i]++]=c,c=getchar();
	}
    for (rr int i=1;i<n;++i){
    	rr int x=iut(),y=iut();
    	e[++et]=(node){y,as[x]},as[x]=et,
    	e[++et]=(node){x,as[y]},as[y]=et; 
	}
	dfs1(1,0); for (S=1;S<N;S<<=1) dfs2(1,0);
	for (rr int i=1;i<=n;++i) print(ans[i]),putchar(10);
	return 0;
}