51nod1656 合并trie树

给每个子树存一个 size 因为 我们只需要每个字符出现一次 所以如果x 或者 y一个没有的话才能返回 不然的话 就会重复

#include<iostream>
#include<cstring>
#define x first
#define y second

using namespace std;

const int N = (3e5 + 10) * 2,M = N * 2;

typedef pair<int,int> PII;
typedef long long ll;

int a[N];char b[N];
int head[N],to[M],last[M],cnt;
void add(int a,int b){
    
    to[++cnt] = b;
    last[cnt] = head[a];
    head[a] = cnt;
}
int tree[N][30];
int v[N],T[N];
int ans;
int insert(int p,char a){
    
    if(!p) p = ++ans;
    ++ans;
    for(int i = 0; i <= 25; i++){
    
        tree[ans][i] = tree[p][i];
        tree[p][i] = 0;
    }
    v[ans] += v[p];
    tree[p][a - 'a'] = ans;
    v[ans]++;
    v[p] = v[ans];
    return p;
}

PII merge(int x,int y){
    
    if(!x) return {
    y,v[y]};
    if(!y) return {
    x,v[x]};
    for(int i = 0; i <= 25; i++){
    
        int d = v[tree[x][i]];
        PII p = merge(tree[x][i],tree[y][i]);
        if(p.x == tree[y][i]) v[x] += p.y;
        if(p.x == tree[x][i]) v[x] -= d,v[x] += p.y;
        tree[x][i] = p.x;
    }
    return {
    x,v[x]};
}

ll maxn = 0,has = 0;

void dfs(int x,int pre){
    
    int now = 0;
    for(int i = head[x]; i != -1; i = last[i]){
    
        int j = to[i];
        if(j == pre) continue;
        dfs(j,x);
        if(!now) now = j;
        else{
    
            PII d = merge(T[now],T[j]);
            T[now] = d.x;
        }
    }
    T[0] = 0;
    T[x] = insert(T[now],b[x]);
    if(maxn < v[T[x]] + a[x]){
    
        maxn = v[T[x]] + a[x];

        has = 1;
    }else if(maxn == v[T[x]] + a[x]) has++;

}

int main(){
    
    int n;
    cin >> n;
    memset(head,-1,sizeof head);
    for(int i = 1; i <= n; i++){
    
        scanf("%d",&a[i]);
    }

    scanf("%s",b + 1);

    for(int i = 1; i <= n - 1; i++){
    
        int x,y;
        scanf("%d%d",&x,&y);
        add(x,y);
        add(y,x);
    }

    dfs(1,0);


    cout << maxn << endl;
    cout << has << endl;


    return 0;
}