[HNOI2014]米特运输

显然知道一个节点就可以推出整棵树
然而直接乘会爆longlong
所以考虑取log
最后排序算众数即可

# include <stdio.h>
# include <stdlib.h>
# include <iostream>
# include <algorithm>
# include <string.h>
# include <math.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL ll Read(){
    RG char c = getchar(); RG ll x = 0, z = 1;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + c - '0';
    return x * z;
}

const int MAXN(500010);
const double EPS(1e-8);
int n, cnt, ft[MAXN], a[MAXN], son[MAXN], ans;
double s[MAXN];
struct Edge{  int to, nt;  } edge[MAXN << 1];
/*
  s[i]表示i不变，根的a取对数后的值
  对s排序，取最多的众数个数num
  ans = n - num
*/

IL void Add(RG int u, RG int v){
    edge[cnt] = (Edge){v, ft[u]}; ft[u] = cnt++; son[u]++;
}

IL void Dfs(RG int u){
    for(RG int e = ft[u]; e != -1; e = edge[e].nt){
        RG int v = edge[e].to;
        s[v] = s[u] + log(son[u]);
        Dfs(v);
    }
    s[u] += log(a[u]);
}

int main(RG int argc, RG char* argv[]){
    n = Read();
    for(RG int i = 1; i <= n; i++) a[i] = Read(), ft[i] = -1;
    for(RG int i = 1, u, v; i < n; i++) u = Read(), v = Read(), Add(u, v);
    Dfs(1);
    sort(s + 1, s + n + 1);
    for(RG int i = 2, num = 1; i <= n; i++)
        if(s[i] - s[i - 1] < EPS) num++, ans = max(ans, num);
        else num = 1;
    printf("%d
", n - ans);
    return 0;
}