【传送门：BZOJ4477】
简要题意：
给出一棵n个点的树，树上的边都代表一个字符串，给出Q个询问，每个询问输入x,y和字符串s，求出x到y的路径上以s为前缀的字符串个数
题解：
自己yy了一波可持久化字典树
将每条边的字符串放到深度更深的节点保存
对于x到y的路径，将rt[x]+rt[y]-2*rt[lca]就可以了，因为我们不需要lca到它父亲的字符串
本机RE，提交AC。。。怕是机子出毛病了
参考代码：
#include<cstdio>#include<cstdlib>#include<algorithm>#include<cmath>#include<cstring>using namespace std;struct trie{ ???int son[27],c; ???trie() ???{ ???????c=0; ???????memset(son,0,sizeof(son)); ???}}tr[4100000];int trlen;int rt[110000];struct node{ ???int x,y,next;}a[110000];int len,last[110000];void ins(int x,int y){ ???len++; ???a[len].x=x;a[len].y=y; ???a[len].next=last[x];last[x]=len;}char st[21];int slen;void Link(int &u1,int d){ ???if(u1==0) u1=++trlen; ???tr[u1].c=1; ???if(d==slen) return ; ???d++; ???int y=st[d]-‘a‘+1; ???Link(tr[u1].son[y],d);}void Merge(int &u1,int u2){ ???if(u1==0){u1=u2;return ;} ???if(u2==0) return ; ???tr[u1].c+=tr[u2].c; ???for(int i=1;i<=26;i++) Merge(tr[u1].son[i],tr[u2].son[i]);}int f[110000][21],bin[21],dep[110000];void dfs(int x){ ???for(int i=1;bin[i]<=dep[x];i++) f[x][i]=f[f[x][i-1]][i-1]; ???for(int k=last[x];k;k=a[k].next) ???{ ???????int y=a[k].y; ???????if(y!=f[x][0]) ???????{ ???????????f[y][0]=x; ???????????dep[y]=dep[x]+1; ???????????Merge(rt[y],rt[x]); ???????????dfs(y); ???????} ???}}int LCA(int x,int y){ ???if(dep[x]<dep[y]) swap(x,y); ???for(int i=20;i>=0;i--) ???{ ???????if(dep[x]-dep[y]>=bin[i]) ???????{ ???????????x=f[x][i]; ???????} ???} ???if(x==y) return x; ???for(int i=20;i>=0;i--) ???{ ???????if(dep[x]>=bin[i]&&f[x][i]!=f[y][i]) ???????{ ???????????x=f[x][i];y=f[y][i]; ???????} ???} ???return f[x][0];}char s[21];int clen;int findc(int u1,int u2,int u3,int d){ ???int c=tr[u1].c+tr[u2].c-2*tr[u3].c; ???if(c==0) return 0; ???if(d==clen) return c; ???d++; ???int y=s[d]-‘a‘+1; ???return findc(tr[u1].son[y],tr[u2].son[y],tr[u3].son[y],d);}int main(){ ???bin[0]=1;for(int i=1;i<=20;i++) bin[i]=bin[i-1]<<1; ???int n; ???scanf("%d",&n); ???len=0;memset(last,0,sizeof(last)); ???trlen=0;memset(rt,0,sizeof(rt)); ???for(int i=1;i<n;i++) ???{ ???????int x,y; ???????scanf("%d%d%s",&x,&y,st+1);slen=strlen(st+1); ???????ins(x,y); ???????Link(rt[y],0); ???} ???f[1][0]=0;dep[1]=0;dfs(1); ???int Q; ???scanf("%d",&Q); ???for(int i=1;i<=Q;i++) ???{ ???????int x,y; ???????scanf("%d%d%s",&x,&y,s+1); ???????clen=strlen(s+1); ???????int lca=LCA(x,y); ???????printf("%d\n",findc(rt[x],rt[y],rt[lca],0)); ???} ???return 0;}BZOJ4477: [Jsoi2015]字符串树
原文地址：https://www.cnblogs.com/Never-mind/p/8926034.html