codeforces 914E Palindromes in a Tree（点分治）

You are given a tree (a connected acyclic undirected graph) of n vertices. Vertices are numbered from 1 to n and each vertex is assigned a character from a to t.

A path in the tree is said to be palindromic if at least one permutation of the labels in the path is a palindrome.

For each vertex, output the number of palindromic paths passing through it.

Note: The path from vertex u to vertex v is considered to be the same as the path from vertex v to vertex u, and this path will be counted only once for each of the vertices it passes through.

Input

The first line contains an integer n (2 ≤ n ≤ 2·105)  — the number of vertices in the tree.

The next n - 1 lines each contain two integers u and v (1  ≤  u, v  ≤  n, u ≠ v) denoting an edge connecting vertex u and vertex v. It is guaranteed that the given graph is a tree.

The next line contains a string consisting of n lowercase characters from a to t where the i-th (1 ≤ i ≤ n) character is the label of vertex i in the tree.

Output

Print n integers in a single line, the i-th of which is the number of palindromic paths passing through vertex i in the tree.

Examples

input

Copy

5
1 2
2 3
3 4
3 5
abcbb

output

Copy

1 3 4 3 3 

input

Copy

7
6 2
4 3
3 7
5 2
7 2
1 4
afefdfs

output

1 4 1 1 2 4 2 

Note

In the first sample case, the following paths are palindromic:

2 - 3 - 4

2 - 3 - 5

4 - 3 - 5

Additionally, all paths containing only one vertex are palindromic. Listed below are a few paths in the first sample that are not palindromic:

1 - 2 - 3

1 - 2 - 3 - 4

1 - 2 - 3 - 5

 

题意：给你一颗 n 个顶点的树（连通无环图）。顶点从 1 到 n 编号，并且每个顶点对应一个在‘a’到‘t’的字母。 树上的一条路径是回文是指至少有一个对应字母的排列为回文。 对于每个顶点，输出通过它的回文路径的数量。 注意：从u到v的路径与从v到u的路径视为相同，只计数一次。

 

题解：

首先是一个结论
本题要求路径的某个排列为回文串，很显然，路径上最多有一个字母出现奇数次。只需要记录奇偶性显然可以用状压去解决。
根据这个结论，我们可以推出对于一个点u，他到某个点v的路径如果满足条件，显然状压的数字等于0或者1<<i(i<=19)，这个时候可以考虑搞出一个中点k，于是问题等价于dis(u,k)^dis(v,k)等于0或者1<<i(i<=19)
这显然是点分治的思路
考虑点分治中的暴力怎么写：首先对于重心进行dfs，求出所有点到重心的dis，dis表示路径上个字母的奇偶性，并且记录 dis_idisi​ 的数量。
接着对于每棵子树将该子树对所有dis的数量贡献全部去掉，再对该子树进行dfs，对于每个点的dis，统计与他异或等于0或者1<<i(i<=19)的数量，即为该点贡献答案，值得注意的是，经过该点的路径一定经过其父节点，因为是从另一颗子树中跑过来的，所以在dfs返回的时候应该给父节点加上子节点的贡献。接着把该子树的dis全部加回去跑下一棵子树。
因为每个点本身都是一个回文串，所以最后答案要加一。

 

代码如下：

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define poi void
#define int long long
using namespace std;

vector<int> g[200010];
char c[200010];
int n,ans[200010],a[200010],size[200010],vis[200010],f[200010],tmp[(1<<20)|10]; 

poi get_size(int now,int fa)
{
    size[now]=1;
    f[now]=fa;
    for(int i=0;i<g[now].size();i++)
    {
        if(vis[g[now][i]]||g[now][i]==fa) continue;
        get_size(g[now][i],now);
        size[now]+=size[g[now][i]];
    }
}

int get_zx(int now,int fa)
{
    if(size[now]==1) return now;
    int son,maxson=-1;
    for(int i=0;i<g[now].size();i++)
    {
        if(vis[g[now][i]]||g[now][i]==fa) continue;
        if(maxson<size[g[now][i]])
        {
            maxson=size[g[now][i]];
            son=g[now][i];
        }
    }
    int zx=get_zx(son,now);
    while(size[zx]<2*(size[now]-size[zx])) zx=f[zx];
    return zx;
}

poi get(int now,int fa,int sta,int kd)
{ 
    sta^=(1<<a[now]);
    tmp[sta]+=kd;
    for(int i=0;i<g[now].size();i++)
    {
        if(vis[g[now][i]]||g[now][i]==fa) continue;
        get(g[now][i],now,sta,kd);
    }
}

int calc(int now,int fa,int sta)
{
    sta^=(1<<a[now]);
    int num=tmp[sta];
    for(int i=0;i<=19;i++)
    {
        num+=tmp[sta^(1<<i)];
    }
    for(int i=0;i<g[now].size();i++)
    {
        if(vis[g[now][i]]||g[now][i]==fa) continue;
        num+=calc(g[now][i],now,sta);
    }
    ans[now]+=num;
    return num;
}

poi solve(int now)
{
    vis[now]=1;
    get(now,0,0,1);
    long long num=tmp[0];
    for(int i=0;i<=19;i++)
    {
        num+=tmp[1<<i];
    }
    for(int i=0;i<g[now].size();i++)
    {
        if(vis[g[now][i]]) continue;
        get(g[now][i],0,1<<a[now],-1);
        num+=calc(g[now][i],0,0);
        get(g[now][i],0,1<<a[now],1);
    }
    ans[now]+=num/2;
    get(now,0,0,-1);
    for(int i=0;i<g[now].size();i++)
    {
        if(vis[g[now][i]]) continue;
        get_size(g[now][i],0);
        int zx=get_zx(g[now][i],0);
        solve(zx);
    }
}

signed main()
{
    scanf("%lld",&n);
    for(int i=1;i<n;i++)
    {
        int from,to;
        scanf("%lld%lld",&from,&to);
        g[from].push_back(to);
        g[to].push_back(from);
    }
    scanf("%s",c+1);
    for(int i=1;i<=n;i++)
    {
        a[i]=c[i]-'a';
    }
    solve(1);
    for(int i=1;i<=n;i++)
    {
        printf("%lld ",ans[i]+1);
    }
}