BZOJ4477 JSOI2015字符串树（可持久化trie）

　　树上建可持久化trie即可，有点过于裸了。darkbzoj过了然而在bzoj一直wa，不知道哪有锅。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 100010
int n,m,p[N],fa[N][19],deep[N],root[N],t,cnt;
struct data{int to,nxt;char s[11];
}edge[N<<1];
struct data2{int x,ch[26];
}tree[N*11];
void addedge(int x,int y,char *s){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;memcpy(edge[t].s,s,sizeof(s));}
void ins(int &k,int p,char *s,int n)
{
    tree[++cnt]=tree[k],k=cnt;tree[k].x++;
    if (p==n) return;
    ins(tree[k].ch[s[p+1]-'a'],p+1,s,n);
}
int query(int k,int p,char *s,int n)
{
    if (!k) return 0;
    if (p==n) return tree[k].x;
    return query(tree[k].ch[s[p+1]-'a'],p+1,s,n);
}
void dfs(int k)
{
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k][0])
    {
        fa[edge[i].to][0]=k;
        deep[edge[i].to]=deep[k]+1;
        root[edge[i].to]=root[k];
        ins(root[edge[i].to],0,edge[i].s,strlen(edge[i].s+1));
        dfs(edge[i].to);
    }
}
int lca(int x,int y)
{
    if (deep[x]<deep[y]) swap(x,y);
    for (int j=18;~j;j--) if (deep[fa[x][j]]>=deep[y]) x=fa[x][j];
    if (x==y) return x;
    for (int j=18;~j;j--) if (fa[x][j]!=fa[y][j]) x=fa[x][j],y=fa[y][j];
    return fa[x][0];
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4477.in","r",stdin);
    freopen("bzoj4477.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=1;i<n;i++)
    {
        int x=read(),y=read();
        char s[11];scanf("%s",s+1);
        addedge(x,y,s);
    }
    dfs(1);
    fa[1][0]=1;
    for (int j=1;j<19;j++)
        for (int i=1;i<=n;i++)
        fa[i][j]=fa[fa[i][j-1]][j-1];
    m=read();
    while (m--)
    {
        int x=read(),y=read();
        char s[11];scanf("%s",s+1);
        printf("%d\n",query(root[x],0,s,strlen(s+1))+query(root[y],0,s,strlen(s+1))-(query(root[lca(x,y)],0,s,strlen(s+1))<<1));
    }
    return 0;
}