题意
Sol
广义SAM的板子题。
首先叶子节点不超过20，那么可以直接对每个叶子节点为根的子树插入到广义SAM中。
因为所有合法的答案一定是某个叶子节点为根的树上的一条链，因此这样可以统计出所有合法的答案
然后就是经典的本质不同子串问题了，(ans = sum len[i] - len[fa[i]])
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN = 2e6 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, C, a[MAXN], deg[MAXN];
namespace SAM {
int fa[MAXN], len[MAXN], ch[MAXN][11], tot = 1, root = 1, las = 1;
int insert(int x, int pre) {
int now = ++tot; len[now] = len[pre] + 1;
for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
if(!pre) {fa[now] = root; return now;}
int q = ch[pre][x];
if(len[q] == len[pre] + 1) fa[now] = q;
else {
int nq = ++tot; fa[nq] = fa[q]; len[nq] = len[pre] + 1;
memcpy(ch[nq], ch[q], sizeof(ch[q]));
for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq;
fa[q] = fa[now] = nq;
}
return now;
}
LL calc() {
LL ans = 0;
for(int i = 1; i <= tot; i++) ans += len[i] - len[fa[i]];
return ans;
}
}
vector<int> v[MAXN], node;
void dfs(int x, int _fa, int p) {
p = SAM::insert(a[x], p);
for(auto &to : v[x]) {
if(to == _fa) continue;
dfs(to, x, p);
}
}
int main() {
N = read(); C = read();
for(int i = 1; i <= N; i++) a[i] = read();
for(int i = 1; i <= N - 1; i++) {
int x = read(), y = read();
v[x].push_back(y); deg[x]++;
v[y].push_back(x); deg[y]++;
}
for(int i = 1; i <= N; i++)
if(deg[i] == 1) SAM::las = 1, dfs(i, 0, 1);
cout << SAM::calc();
return 0;
}