Educational Codeforces Round 2 E. Lomsat gelral 启发式合并map

E. Lomsat gelral

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/600/problem/E

Description

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.

Sample Input

4
1 2 3 4
1 2
2 3
2 4

Sample Output

10 9 3 4

HINT

题意

给你一棵树，告诉你每个节点的颜色，然后让你输出对于这个节点的子树中，出现次数最多的颜色的权值和是多少

题解：

启发式合并map

对于每一个子树，我们都维护一个map，然后从小的合并到大的中

均摊下来复杂度不会很高（雾

代码：

#include<iostream>
#include<stdio.h>
#include<vector>
#include<map>
using namespace std;
#define maxn 800005

map<int,int> H[maxn];
map<int,int>::iterator it;
long long ans[maxn];
vector<int> E[maxn];
int c[maxn];
int id[maxn];
long long M[maxn];
long long M1[maxn];
void uni(int &x,int y)
{
    if(H[x].size()<H[y].size())swap(x,y);
    for(it = H[y].begin();it!=H[y].end();it++)
    {
        H[x][it->first]+=it->second;
        if(M1[x]==H[x][it->first])
            M[x]+=it->first;
        if(M1[x]<H[x][it->first])
        {
            M1[x]=H[x][it->first];
            M[x]=it->first;
        }
    }
}
void solve(int x,int fa)
{
    H[x][c[x]]=;
    M1[x]=,M[x]=c[x];
    for(int i=;i<E[x].size();i++)
    {
        if(E[x][i]==fa)continue;
        solve(E[x][i],x);
        uni(id[x],id[E[x][i]]);
    }
    ans[x]=M[id[x]];
}
long long flag = ;
int main()
{
    int n;scanf("%d",&n);
    for(int i=;i<=n;i++)
    {
        id[i]=i;
        scanf("%d",&c[i]);
    }
    for(int i=;i<n;i++)
    {
        int x,y;scanf("%d%d",&x,&y);
        E[x].push_back(y);
        E[y].push_back(x);
    }
    solve(,-);
    for(int i=;i<=n;i++)
        printf("%lld ",ans[i]);
    printf("\n");
    return ;
}