Codeforces 600E Lomsat gelral (DSU on Tree)

E. Lomsat gelral

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.

Let's call colour c dominating in the subtree of vertex v if there are no other colours that appear in the subtree of vertex v more times than colour c. So it's possible that two or more colours will be dominating in the subtree of some vertex.

The subtree of vertex v is the vertex v and all other vertices that contains vertex v in each path to the root.

For each vertex v find the sum of all dominating colours in the subtree of vertex v.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of vertices in the tree.

The second line contains n integers ci (1 ≤ ci ≤ n), ci — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the edge of the tree. The first vertex is the root of the tree.

Output

Print n integers — the sums of dominating colours for each vertex.

Examples

input

41 2 3 41 22 32 4

output

10 9 3 4

input

151 2 3 1 2 3 3 1 1 3 2 2 1 2 31 21 31 41 141 152 52 62 73 83 93 104 114 124 13

output

6 5 4 3 2 3 3 1 1 3 2 2 1 2 3

一棵树，每个节点有一个颜色c，现在问你一个节点的子树中，所有出现次数最多的颜色c的数字之和。

树上启发式合并模板题。

写法：链剖写法

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <deque>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <stack>#include <iomanip>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;typedef double db;const int maxn=500005,inf=0x3f3f3f3f;  const ll llinf=0x3f3f3f3f3f3f3f3f;   const ld pi=acos(-1.0L);int c[maxn],cnt[maxn],size[maxn],son[maxn],head[maxn];ll ans[maxn];bool visit[maxn],big[maxn];ll sum,mx,num;struct Edge {int from,to,pre;};Edge edge[maxn*2];void addedge(int from, int to) {edge[num]=(Edge){from,to,head[from]};head[from]=num++;edge[num]=(Edge){to,from,head[to]};head[to]=num++;}int dfs2(int now) {      visit[now]=1;son[now]=-1;size[now]=1;      for (int i=head[now];i!=-1;i=edge[i].pre) {          int to=edge[i].to;          if (!visit[to]) {              size[now]+=dfs2(to);              if (son[now]==-1||size[to]>size[son[now]]) son[now]=to;          }      }    if (son[now]!=-1) big[son[now]]=1;    return size[now];}  void add(int now,int fa,int val) {cnt[c[now]]+=val;if (cnt[c[now]]>mx) {mx=cnt[c[now]];sum=c[now];} else if (cnt[c[now]]==mx) sum+=(ll)c[now];for (int i=head[now];i!=-1;i=edge[i].pre) {int to=edge[i].to;if (to!=fa&&!big[to]) add(to,now,val);}}void dfs(int now,int fa,bool rem) {visit[now]=1;for (int i=head[now];i!=-1;i=edge[i].pre) {int to=edge[i].to;if (to!=fa&&!big[to]) dfs(to,now,0);}if (son[now]!=-1) dfs(son[now],now,1);add(now,fa,1);ans[now]=sum;if (son[now]!=-1) big[son[now]]=0;if (!rem) add(now,fa,-1),sum=mx=0;}int main() {int n,i,j,x,y;num=0;memset(head,-1,sizeof(head));scanf("%d",&n);for (i=1;i<=n;i++) {scanf("%d",&c[i]);}for (i=1;i<n;i++) {scanf("%d%d",&x,&y);addedge(x,y);}mem0(visit);mem0(big);dfs2(1);mem0(visit);sum=mx=0;dfs(1,0,0);for (i=1;i<=n;i++) printf("%I64d ",ans[i]);return 0;}