給定一棵 n個節點的樹，每個節點有一個小寫字母。
有 m 組詢問，每組詢問為樹上 a→b 和 c→d 組成的字串的最長公共字首。
\(n≤3×10^5，m≤10^6\)。
兩個字串求任意子串的最長公共字首，可以二分+雜湊。
樹上的路徑透過樹剖分出\(log(n)\)個區間得到\(log(n)\)個字串，每次比較一個區間的雜湊值。當區間雜湊值不一樣，二分。
實現要注意很多細節。a→lca, lca→b 要分開考慮並且求雜湊。
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define P pair<LL, LL>
#define mk make_pair
#define fs first
#define sc second
const int N = 3e5 + 11;
const LL modd[11] = {19260817, 381335473, 252412907, 173275057, 415924151, 490280677, 215695621, 311094167};
bool vis[11];
LL mod1, mod2;
int n, m;
int head[N], nex[N<<1], to[N<<1], size;
int dep[N], sz[N], dfn[N], son[N], id[N], fa[N], top[N], cnt;
char s[N];
P B = mk(127, 23333), h1[N], h2[N], p[N];
vector<P> up, dw, f, g;
P operator * (P a, P b){
return mk(a.fs * b.fs % mod1, a.sc * b.sc % mod2);
}
P operator + (P a, P b){
return mk((a.fs + b.fs) % mod1, (a.sc + b.sc) % mod2);
}
P operator - (P a, P b){
return mk((a.fs + mod1 - b.fs) % mod1, (a.sc + mod2 - b.sc) % mod2);
}
int get_mod(){
int x;
do{
x = rand() % 8;
}while(vis[x]);
vis[x] = 1;
return modd[x];
}
int read(){
int x = 0;
char ch = getchar();
while(ch > '9' || ch < '0')ch = getchar();
while(ch >= '0' && ch <= '9')x = 10 * x + ch - 48, ch = getchar();
return x;
}
void print(int x){
if(x > 9)print(x / 10);
putchar(x % 10 + 48);
}
void dfs(int u){
sz[u] = 1;
dep[u] = dep[fa[u]] + 1;
for(int i = head[u];i;i = nex[i]){
int v = to[i];
if(v == fa[u])continue;
fa[v] = u;
dfs(v);
sz[u] += sz[v];
if(sz[son[u]] < sz[v])son[u] = v;
}
}
void dfs(int u, int tp){
top[u] = tp;
dfn[u] = ++cnt;
id[cnt] = u;
if(son[u])dfs(son[u], tp);
for(int i = head[u];i; i = nex[i]){
int v = to[i];
if(v == fa[u] || v == son[u])continue;
dfs(v, v);
}
}
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] ? y : x;
}
vector<P> get(int x, int y){
int lca = LCA(x, y);
up.clear();
while(top[x] != top[lca])up.push_back(mk(dfn[x], dfn[top[x]])), x = fa[top[x]];
up.push_back(mk(dfn[x], dfn[lca]));
while(top[y] != top[lca])dw.push_back(mk(dfn[top[y]], dfn[y])), y = fa[top[y]];
if(lca != y){
dw.push_back(mk(dfn[son[lca]], dfn[y]));
}
while(dw.size()) up.push_back(dw.back()), dw.pop_back();
return up;
}
P get(int fl, int st, int len){
if(fl)return h2[st-len+1] - h2[st+1] * p[len];
return h1[st+len-1] - h1[st-1] * p[len];
}
void add(int x, int y){
to[++size] = y; nex[size] = head[x]; head[x] = size;
}
int main(){
srand(time(0));
mod1 = get_mod();
mod2 = get_mod();
cin>>n;
scanf("%s", s + 1);
p[0] = mk(1, 1); p[1] = B;
int u, v;
for(int i = 2;i <= n; i++){
u = read(); v = read();
add(u, v);add(v, u);
p[i] = p[i-1] * B;
}
dfs(1);
dfs(1, 1);
for(int i = 1;i <= n; i++){
int x = s[id[i]];
h1[i] = h1[i-1] * B + mk(x, x);
}
for(int i = n;i >= 1; i--){
int x = s[id[i]];
h2[i] = h2[i+1] * B + mk(x, x);
}
m = read();
while(m--){
int a = read(), b = read(), c = read(), d = read();
f = get(a, b); g = get(c, d);
int i = 0, j = 0, ans = 0;
while(i < f.size() && j < g.size()){
int df1 = f[i].fs, df2 = f[i].sc;
int dg1 = g[j].fs, dg2 = g[j].sc;
int flf = df2 < df1, flg = dg2 < dg1;
int lf = abs(df1 - df2) + 1, lg = abs(dg2 - dg1) + 1;
int len = min(lf, lg);
P hf = get(flf, df1, len), hg = get(flg, dg1, len);
if(hf == hg){
if(len == lf)i++;
else f[i].fs = flf == 1 ? df1 - len : df1 + len;
if(len == lg)j++;
else g[j].fs = flg == 1 ? dg1 - len : dg1 + len;
ans += len;
}
else{
int l = 0, r = len, res = 0;
while(l <= r){
int mid = l + r >> 1;
if(get(flf, df1, mid) == get(flg, dg1, mid))l = mid + 1, res = mid;
else r = mid - 1;
}
ans += res;
break;
}
}
print(ans);
puts("");
}
return 0;
}