题解
按秩合并怎么清数组对我来说真是世纪性难题
我们很熟练地想到点分，如果我们认为某个点到重心是正着读的，由于它的深度固定，它的串也是固定的，我们只要预处理出所有长度正着重复的串，反着重复的串，和它们的哈希值，遍历树的时候只需要记录一下路径字符串的哈希值，比对一下看是否合法就行
为了快一点可以按深度合并每棵子树
代码
#include <bits/stdc++.h>
#define enter putchar('
')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define MAXN 1000005
#define mo 999999137
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
char s[MAXN],str1[MAXN],str2[MAXN],s1[MAXN],s2[MAXN];
int e[MAXN],h1[MAXN],h2[MAXN];
struct node {
int to,next;
}E[MAXN * 2];
int head[MAXN],sumE,N,M;
int que[MAXN],ql,qr,fa[MAXN],siz[MAXN],son[MAXN],H[MAXN],dep[MAXN];
int t1[MAXN],t2[MAXN],sum1[MAXN],sum2[MAXN];
int64 ans;
bool vis[MAXN];
vector<pii > ver;
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
int mul(int a,int b) {
return 1LL * a * b % mo;
}
int inc(int a,int b) {
return a + b >= mo ? a + b - mo : a + b;
}
void Init() {
memset(head,0,sizeof(head));sumE = 0;
memset(vis,0,sizeof(vis));ans = 0;
read(N);read(M);
scanf("%s",s + 1);
int u,v;
for(int i = 1 ; i < N ; ++i) {
read(u);read(v);
add(u,v);add(v,u);
}
scanf("%s",str1 + 1);
for(int i = 1 ; i <= N ; ++i) {
s1[i] = str1[(i - 1) % M + 1];
}
memcpy(str2,str1,sizeof(str1));
reverse(str2 + 1,str2 + M + 1);
for(int i = 1 ; i <= N ; ++i) {
s2[i] = str2[(i - 1) % M + 1];
}
for(int i = 1 ; i <= N ; ++i) {
h1[i] = inc(mul(e[i - 1],s1[i] - 'A' + 1),h1[i - 1]);
h2[i] = inc(mul(e[i - 1],s2[i] - 'A' + 1),h2[i - 1]);
}
}
int CalcG(int st) {
fa[st] = 0;que[ql = qr = 1] = st;
while(ql <= qr) {
int u = que[ql++];siz[u] = 1;son[u] = 0;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v] && v != fa[u]) {
que[++qr] = v;
fa[v] = u;
}
}
}
int res = que[qr];
for(int i = qr ; i >= 1 ; --i) {
int u = que[i];
if(son[u] < qr - siz[u]) son[u] = qr - siz[u];
if(fa[u]) {
siz[fa[u]] += siz[u];
son[fa[u]] = max(son[fa[u]],siz[u]);
}
if(son[u] < son[res]) res = u;
}
return res;
}
void Calc(int st) {
que[ql = qr = 1] = st;H[st] = s[st] - 'A' + 1;dep[st] = 1;
fa[st] = 0;
while(ql <= qr) {
int u = que[ql++];
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa[u] && !vis[v]) {
fa[v] = u;
H[v] = inc(mul(H[u],e[1]),s[v] - 'A' + 1);
dep[v] = dep[u] + 1;
que[++qr] = v;
}
}
}
for(int i = 1 ; i <= qr ; ++i) {
int u = que[i];
if(H[u] == h1[dep[u]]) t1[dep[u] % M]++;
if(H[u] == h2[dep[u]]) t2[dep[u] % M]++;
}
}
int Get_Dep(int u,int fa) {
int res = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v] && v != fa) {
res = max(res,1 + Get_Dep(v,u));
}
}
return res;
}
void dfs(int u) {
int G = CalcG(u);
vis[G] = 1;
ver.clear();
H[G] = 0;
for(int i = head[G] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) ver.pb(mp(Get_Dep(v,G),v));
}
sort(ver.begin(),ver.end());
int siz = ver.size();
int t = min(ver[siz - 1].fi,M);
for(int i = 0 ; i <= t ; ++i) sum1[i] = sum2[i] = 0;
t = 0;sum1[0] = sum2[0] = 1;
for(int i = 0 ; i < siz ; ++i) {
pii p = ver[i];
for(int j = 0 ; j <= p.fi ; ++j) t1[j] = t2[j] = 0;
Calc(p.se);
for(int j = 0 ; j <= t ; ++j) {
if(M - 1 - j > p.fi) continue;
if(s[G] == str1[j + 1]) ans = inc(ans,mul(sum1[j],t2[M - 1 - j]));
if(s[G] == str2[j + 1]) ans = inc(ans,mul(sum2[j],t1[M - 1 - j]));
}
for(int j = 0 ; j <= p.fi ; ++j) {sum1[j] += t1[j];sum2[j] += t2[j];}
t = min(M - 1,p.fi);
}
for(int i = head[G] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) dfs(v);
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
int T;
read(T);
e[0] = 1;
for(int i = 1 ; i <= 1000000 ; ++i) e[i] = 1LL * e[i - 1] * 47 % mo;
while(T--) {
Init();
dfs(1);
out(ans);enter;
}
return 0;
}