题面
Sol
求多个串的不同子串的个数
广义后缀自动机
也就是可以表示所有的串的所有的后缀的自动机
那么每次建一个串后，另外一个串接在初始节点下面建就好了
叶子节点最多(20)个
那么对于每个叶子结点遍历一遍树建立(sam)
注意回溯时要把(last)指回来
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(1e5 + 5);
const int maxm(4e6 + 5);
int trans[10][maxm], fa[maxm], tot = 1, last, len[maxm];
int n, first[maxn], cnt, col[maxn], degree[maxn];
ll ans;
struct Edge{
int to, next;
} edge[maxn << 1];
IL void Add(RG int u, RG int v){
edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;
}
IL void Extend(RG int c, RG int p){
RG int np = ++tot;
len[last = np] = len[p] + 1;
while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
if(!p) fa[np] = 1;
else{
RG int q = trans[c][p];
if(len[q] == len[p] + 1) fa[np] = q;
else{
RG int nq = ++tot;
fa[nq] = fa[q], len[nq] = len[p] + 1;
for(RG int i = 0; i < 10; ++i) trans[i][nq] = trans[i][q];
fa[q] = fa[np] = nq;
while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
}
}
}
IL void Dfs(RG int u, RG int ff, RG int p){
Extend(col[u], p), p = last;
for(RG int e = first[u]; e != -1; e = edge[e].next)
if(edge[e].to != ff) Dfs(edge[e].to, u, p);
}
int main(RG int argc, RG char *argv[]){
n = Input(), Input();
for(RG int i = 1; i <= n; ++i) first[i] = -1, col[i] = Input();
for(RG int i = 1; i < n; ++i){
RG int u = Input(), v = Input();
Add(u, v), Add(v, u), ++degree[u], ++degree[v];
}
for(RG int i = 1; i <= n; ++i)
if(degree[i] == 1) last = 1, Dfs(i, 0, 1);
for(RG int i = 1; i <= tot; ++i) ans += len[i] - len[fa[i]];
printf("%lld
", ans);
return 0;
}