Lomsat gelral CodeForces - 600E（树上启发式合并）
Lomsat gelral CodeForces - 600E
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 c i (1 ≤ c i ≤ n), c i — the colour of the i-th vertex.
Each of the next n - 1 lines contains two integers x j, y j (1 ≤ x j, y j ≤ 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
4
1 2 3 4
1 2
2 3
2 4
Output
10 9 3 4
Input
15
1 2 3 1 2 3 3 1 1 3 2 2 1 2 3
1 2
1 3
1 4
1 14
1 15
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13
Output
6 5 4 3 2 3 3 1 1 3 2 2 1 2 3
#include<bits/stdc++.h>
using namespace std;
const long long maxn = 1e5+9;
long long cnt[maxn],ans[maxn],val[maxn],sum,son[maxn],Size[maxn],flag[maxn],max_cnt;
vector<long long>edge[maxn];
void read(long long &x)
{
x = 0;
long long f = 1;
char ch = getchar();
while(!isdigit(ch))
{
if(ch == '-')f = -1;
ch = getchar();
}
while(isdigit(ch))
{
x = (x<<3) + (x<<1) + ch - '0';
ch = getchar();
}
x = x*f;
}
void con_son(long long x,long long fa)
{
long long max_num = -1;
Size[x]=1;
for(long long i = 0; i<edge[x].size(); i++)
{
long long to=edge[x][i];
if(to==fa)
continue;
con_son(to,x);
if(Size[to]>max_num)
{
max_num = Size[to];
son[x] = to;
}
Size[x]+=Size[to];
}
}
void conclute(long long x,long long fa,long long add)
{
cnt[val[x]]+=add;
if(add==1&&cnt[val[x]]>=max_cnt)
{
if(cnt[val[x]]>max_cnt)
sum = val[x];
else
sum += val[x];
max_cnt = cnt[val[x]];
}
for(long long i=0; i<edge[x].size(); i++)
{
long long to=edge[x][i];
if(to==fa||flag[to])
continue;
conclute(to,x,add);
}
}
void dfs(long long x,long long fa,long long save)
{
for(long long i = 0; i<edge[x].size(); i++)
{
long long to=edge[x][i];
if(to==fa||to==son[x])
continue;
dfs(to,x,0);
}
if(son[x])
{
dfs(son[x],x,1);
flag[son[x]]=1;
}
conclute(x,fa,1);
ans[x]=sum;
if(son[x])
flag[son[x]]=0;
if(save==0)
{
conclute(x,fa,-1);
sum=max_cnt=0;
}
}
int main()
{
long long i,j,m,n,a,b;
ios::sync_with_stdio(false);
read(n);
for(i = 1; i<=n; i++)
{
read(val[i]);
}
for(i =1; i<=n-1; i++)
{
read(a);
read(b);
edge[a].push_back(b);
edge[b].push_back(a);
}
con_son(1,0);
dfs(1,0,1);
for(i = 1; i<=n; i++)
{
printf("%lld%c",ans[i],i==n?'\n':' ');
}
return 0;
}