BZOJ3573 HNOI2014米特运输

　　显然确定一个点的权值后整棵树权值确定。只要算出根节点的权值就能知道两种改法是否等价。

　　乘的话显然会炸，取log即可。map似乎会出一些问题，sort即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 500010
#define double long double
const double eps=1E-;
int n,a[N],p[N],degree[N],t=;
double f[N];
struct data{int to,nxt;
}edge[N<<];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dfs(int k,int from)
{
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from) degree[k]++;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from)
    {
        f[edge[i].to]=f[k]+log(degree[k]);
        dfs(edge[i].to,k);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3573.in","r",stdin);
    freopen("bzoj3573.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<=n;i++) a[i]=read();
    for (int i=;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    dfs(,);
    int ans=;
    for (int i=;i<=n;i++)
    f[i]+=log(a[i]);
    sort(f+,f+n+);
    for (int i=;i<=n;i++)
    {
        int t=i;
        while (t<n&&fabs(f[t+]-f[i])<eps) t++;
        ans=max(ans,t-i+);
        i=t;
    }
    cout<<n-ans;
    return ;
}