Apple Tree CodeForces - 349D

http://codeforces.com/problemset/problem/349/D

对于u来说 要使其子节点v1 v2...vk的苹果数都相等 肯定不能简单把所有节点都置为最小值 因为对于u的子节点也要满足子树苹果数相等 假设v1 v2...vk这几棵子树每次最少要拿掉多少苹果才能使自己的子树平衡是已知的 显然u一次拿多少就是v1 v2...vk的lcm了

因为要使每个节点都满足所有子树相等 所以自底向上考虑即可

#include <bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
const ll N=0x3f3f3f3f3f3f3f3f;
const int maxn=1e5+10;

struct node
{
    int v,next;
};

node edge[2*maxn];
ll sum[maxn],lcmval[maxn];
int first[maxn];
ll ans;
int n,num;

void addedge(int u,int v)
{
    edge[num].v=v;
    edge[num].next=first[u];
    first[u]=num++;
}

ll getlcm(ll a,ll b)
{
    return (a/__gcd(a,b))*b;
}

void dfs(int cur,int fa)
{
    ll minn,tot,sz;
    int i,v;
    lcmval[cur]=1;
    sz=0;
    for(i=first[cur];i!=-1;i=edge[i].next){
        v=edge[i].v;
        if(v!=fa){
            sz++;
            dfs(v,cur);
            if(lcmval[v]!=0) lcmval[cur]=getlcm(lcmval[cur],lcmval[v]);
        }
    }
    if(sz==0) return;
    minn=N,tot=0;
    for(i=first[cur];i!=-1;i=edge[i].next){
        v=edge[i].v;
        if(v!=fa){
            minn=min(minn,sum[v]-sum[v]%lcmval[cur]);
            tot+=sum[v];
        }
    }
    sum[cur]=sz*minn,lcmval[cur]*=sz;
    ans+=tot-sum[cur];
}

int main()
{
    int i,u,v;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%lld",&sum[i]);
    }
    memset(first,-1,sizeof(first));
    num=0;
    for(i=1;i<=n-1;i++){
        scanf("%d%d",&u,&v);
        addedge(u,v),addedge(v,u);
    }
    dfs(1,0);
    printf("%lld\n",ans);
    return 0;
}