BZOJ 3727 DP？推式子..

 思路：

 

 

设$sum[i]表示i的子树中a[i]的和$

$b[1]=\Sigma a[i]*dis[i] = \Sigma _{i=2} ^n sum[i]$

$b[x]-b[fa[x]]=sum[1]-2*sum[x]$

$sum[1]={\Sigma_{i=2}^n (b[x]-b[fa[x]])+2*b[1] \over n-1}$

$求出sum[1]以后根据a[x]=sum[x]-\Sigma_{v是x的儿子}　sum[v]带入求出其它值即可$

$复杂度O(n)$

//By SiriusRen
#include <cstdio>
#include <cstring>
using namespace std;
#define int long long
const int N=600500;
int n,xx,yy,first[N],next[N],v[N],tot,fa[N],rev[N],cnt,b[N],X;
long long sum[N],ans[N];
void add(int x,int y){v[tot]=y,next[tot]=first[x],first[x]=tot++;}
void dfs(int x){
    rev[++cnt]=x;
    for(int i=first[x];~i;i=next[i])if(v[i]!=fa[x])
        fa[v[i]]=x,dfs(v[i]);
}
void dfs2(int x){
    ans[x]=sum[x];
    for(int i=first[x];~i;i=next[i])if(v[i]!=fa[x])
        dfs2(v[i]),ans[x]-=sum[v[i]];
}
signed main(){
    memset(first,-1,sizeof(first));
    scanf("%lld",&n);
    for(int i=1;i<n;i++)scanf("%lld%lld",&xx,&yy),add(xx,yy),add(yy,xx);
    for(int i=1;i<=n;i++)scanf("%lld",&b[i]);
    dfs(1);
    for(int i=2;i<=n;i++)sum[1]+=(b[i]-b[fa[i]]);
    sum[1]=(sum[1]+2*b[1])/(n-1);
    for(int i=2;i<=n;i++)X=rev[i],sum[X]=(sum[1]-b[X]+b[fa[X]])/2;
    dfs2(1);
    for(int i=1;i<=n;i++)printf("%lld%c",ans[i],i!=n?' ':'\n');
}