Educational Codeforces Round 2 E.Lomsat gelral(树形dp)

Educational Codeforces Round 2E：http://codeforces.com/contest/600/problem/E

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.

Sample test(s)

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

题目大意：给定一颗树，节点i的控制颜色为包括该节点在内的所有子树节点中颜色出现最多的（可能有多个控制颜色），求每个节点控制颜色之和

大致思路：从叶子节点向上更新控制颜色即可，但是必须用map映射而且要将小树合并到大树当中，否则会MLE

#include <cstdio>#include <cstring>#include <vector>#include <map>#include <algorithm>using namespace std;int n,c[100005],index[100005],mx[100005];//c[i]表示i节点的颜色，index[i]表示i节点所对应的cnt下标，mx[index[i]]表示i节点相同颜色出现最多的次数long long ans[100005],tmp[100005];//tmp[index[i]]表示i节点当前控制颜色总和map<int,int> cnt[100005];//cnt[index[i]]表示i节点cnt[index[i]].first出现的次数为cnt[index[i]].secondvector<int> v[100005];void Merge(int s,int e) {    if(cnt[index[s]].size()<cnt[index[e]].size())//总是将小树合并到大树上，以防MLE        swap(index[s],index[e]);    for(map<int,int>::iterator it=cnt[index[e]].begin();it!=cnt[index[e]].end();++it) {//枚举小树的map        cnt[index[s]][it->first]+=it->second;//i节点中颜色it->first出现次数加it-second        if(cnt[index[s]][it->first]>mx[index[s]]) {//如果当前颜色it->first出现次数已大于控制颜色出现次数            mx[index[s]]=cnt[index[s]][it->first];//更新控制线色出现次数            tmp[index[s]]=it->first;//重置控制颜色        }        else if(cnt[index[s]][it->first]==mx[index[s]])//如果当前颜色it->first出现次数等于控制颜色出现次数            tmp[index[s]]+=it->first;//控制颜色加上当前颜色    }}void dfs(int pre,int u) {    for(int i=0;i<v[u].size();++i)        if(v[u][i]!=pre) {            dfs(u,v[u][i]);            Merge(u,v[u][i]);        }    ans[u]=tmp[index[u]];}int main() {    int s,e,i;    while(1==scanf("%d",&n)) {        for(i=1;i<=n;++i) {            scanf("%d",c+i);            index[i]=i;            mx[i]=cnt[i][c[i]]=1;            tmp[i]=c[i];        }        for(i=1;i<n;++i) {            scanf("%d%d",&s,&e);            v[s].push_back(e);            v[e].push_back(s);        }        dfs(1,1);        for(i=1;i<n;++i)            printf("%I64d ",ans[i]);        printf("%I64d\n",ans[n]);    }    return 0;}