trie合并的裸题...因为最多只有n个点，所以最多合并n次，复杂度$O(N*26)$。
#include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> using namespace std; const int maxn=500010, inf=1e9; struct poi{int too, pre;}e[maxn<<1]; struct tjm{int nxt[26], size;}tree[maxn]; int n, x, y, ans, cnt, tot; int last[maxn], c[maxn]; char s[maxn]; inline void read(int &k) { int f=1; k=0; char c=getchar(); while(c<'0' || c>'9') c=='-' && (f=-1), c=getchar(); while(c<='9' && c>='0') k=k*10+c-'0', c=getchar(); k*=f; } inline void add(int x, int y){e[++tot]=(poi){y, last[x]}; last[x]=tot;} void merge(int &x, int y) { if(!x || !y) {x+=y; return;} tree[x].size=1; for(int i=0;i<26;i++) merge(tree[x].nxt[i], tree[y].nxt[i]), tree[x].size+=tree[tree[x].nxt[i]].size; } void dfs(int x, int fa) { tree[x].size=1; for(int i=last[x], too;i;i=e[i].pre) if((too=e[i].too)!=fa) { dfs(too, x); int tmp=tree[tree[x].nxt[s[e[i].too]-'a']].size; merge(tree[x].nxt[s[e[i].too]-'a'], e[i].too); tree[x].size+=tree[tree[x].nxt[s[e[i].too]-'a']].size-tmp; } if(tree[x].size+c[x]>ans) ans=tree[x].size+c[x], cnt=1; else if(tree[x].size+c[x]==ans) cnt++; } int main() { read(n); for(int i=1;i<=n;i++) read(c[i]); scanf("%s", s+1); for(int i=1;i<n;i++) read(x), read(y), add(x, y), add(y, x); dfs(1, 0); printf("%d\n%d", ans, cnt); }