解题：CF1009 Dominant Indices

题面

长链剖分模板题

只能按深度统计，同时比DSU on tree难理解一些，但是复杂度少个log

对每个点抓出向下延伸最长的儿子叫做长儿子。在合并时用指针继承信息，对于长儿子O(1)继承，其他儿子暴力

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=;
 int len[N],imp[N],ans[N];
 int p[N],noww[*N],goal[*N];
 int mem[N],*pts[N],*fpt=mem+;
 int n,t1,t2,cnt;
 void Link(int f,int t)
 {
     noww[++cnt]=p[f];
     goal[cnt]=t,p[f]=cnt;
 }
 void DFS(int nde,int fth,int dth)
 {
     for(int i=p[nde];i;i=noww[i])
         if(goal[i]!=fth)
         {
             DFS(goal[i],nde,dth+);
             if(len[goal[i]]>len[imp[nde]])
                 imp[nde]=goal[i];
         }
     len[nde]=len[imp[nde]]+;
 }
 void Getans(int nde,int fth)
 {
     pts[nde][]=;
     if(imp[nde])
     {
         pts[imp[nde]]=pts[nde]+;
         Getans(imp[nde],nde),ans[nde]=ans[imp[nde]]+;
     }
     for(int i=p[nde];i;i=noww[i])
         if(goal[i]!=fth&&goal[i]!=imp[nde])
         {
             pts[goal[i]]=fpt,fpt+=len[goal[i]],Getans(goal[i],nde);
             for(int j=;j<=len[goal[i]];j++)
             {
                 pts[nde][j]+=pts[goal[i]][j-];
                 int b1=(j<=ans[nde]&&pts[nde][j]>=pts[nde][ans[nde]]);
                 int b2=(j>ans[nde]&&pts[nde][j]>pts[nde][ans[nde]]);
                 if(b1||b2) ans[nde]=j;
             }
         }
     if(pts[nde][ans[nde]]==) ans[nde]=;
 }
 int main()
 {
     scanf("%d",&n);
     for(int i=;i<n;i++)
     {
         scanf("%d%d",&t1,&t2);
         Link(t1,t2),Link(t2,t1);
     }
     DFS(,,),pts[]=fpt,fpt+=len[];
     Getans(,);
     for(int i=;i<=n;i++)
         printf("%d\n",ans[i]);
     return ;
 }