Problem:
Score : 500 points

Problem Statement

You are given a tree with N vertices 1,2,\ldots,N, and positive integers c_1,c_2,\ldots,c_N.
The i-th edge in the tree (1 \leq i \leq N-1) connects Vertex a_i and Vertex b_i.

We will write a positive integer on each vertex in T and calculate our score as follows:

On each edge, write the smaller of the integers written on the two endpoints.

Let our score be the sum of the integers written on all the edges.

Find the maximum possible score when we write each of c_1,c_2,\ldots,c_N on one vertex in T, and show one way to achieve it. If an integer occurs multiple times in c_1,c_2,\ldots,c_N, we must use it that number of times.

Constraints

1 \leq N \leq 10000

1 \leq a_i,b_i \leq N

1 \leq c_i \leq 10^5

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}
c_1 \ldots c_N

Output

Use the following format:

M
d_1 \ldots d_N

where M is the maximum possible score, and d_i is the integer to write on Vertex i.
d_1,d_2,\ldots,d_N must be a permutation of c_1,c_2,\ldots,c_N.
If there are multiple ways to achieve the maximum score, any of them will be accepted.

Sample Input 1

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

Sample Output 1

10
1 2 3 4 5

If we write 1,2,3,4,5 on Vertex 1,2,3,4,5, respectively, the integers written on the four edges will be 1,2,3,4, for the score of 10. This is the maximum possible score.

Sample Input 2

5
1 2
1 3
1 4
1 5
3141 59 26 53 59

Sample Output 2

197
59 26 3141 59 53

c_1,c_2,\ldots,c_N may not be pairwise distinct.

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
using System.IO;
using System.Runtime.CompilerServices;

public class ABC
{
    //          long[] sp = Console.ReadLine().Split().Select(long .Parse).ToArray();
    //          int[] sp = Console.ReadLine().Split().Select(int.Parse).ToArray();
    //          int N =  int.Parse(Console.ReadLine());
    //CompLib.Collections.PriorityQueue<?>
    public static int juuokunana = (int)Math.Pow(10, 9) + 7;

    public static void Main()
    {
        int N;
        new readint(out N);

        var nodes = new Tuple<int, int>[N-1];
        Tuple<int, int>[] jisuu = new Tuple<int, int>[N];

        for (int i = 0; i < N; i++)
        {
            jisuu[i] = new Tuple<int, int>(i+1, 0);
        }

        for (int i = 0; i < N-1; i++)
        {
            int[] split = Console.ReadLine().Split().Select(int.Parse).ToArray();
            nodes[i] = new Tuple<int, int>(split[0], split[1]);
            jisuu[split[0]-1] = new Tuple<int, int>(jisuu[split[0]-1].Item1, jisuu[split[0]-1].Item2 + 1);
            jisuu[split[1]-1] = new Tuple<int, int>(jisuu[split[1]-1].Item1, jisuu[split[1]-1].Item2 + 1);

        }

        int[] c = Console.ReadLine().Split().Select(int.Parse).ToArray();

        Array.Sort(jisuu, (a, b) => a.Item2.CompareTo(b.Item2));
        Array.Sort(c);
        Array.Reverse(c);
        int[] assign = new int[N + 1];
        for (int i = 1; i < N+1; i++)
        {
            assign[i] = c[i - 1];
        }
        long score = 0;
        for (int i = 0; i < nodes.Length; i++)
        {
            int a = nodes[i].Item1;
            int b = nodes[i].Item2;//1-base

            score += Math.Min(assign[a], assign[b]);
        }

    }
}


class readint
{



    public readint(out int i)
    {
        i = int.Parse(Console.ReadLine());

    }
    public readint(out int a, out int b)
    {
        int[] sp = Console.ReadLine().Split().Select(int.Parse).ToArray();
        a = sp[0];
        b = sp[1];

    }
    public readint(out int a, out int b, out int c)
    {
        int[] sp = Console.ReadLine().Split().Select(int.Parse).ToArray();
        a = sp[0];
        b = sp[1];
        c = sp[2];
    }
    public readint(out int a, out int b, out int c, out int d)
    {
        int[] sp = Console.ReadLine().Split().Select(int.Parse).ToArray();
        a = sp[0];
        b = sp[1];
        c = sp[2];
        d = sp[3];

    }

}

class readlong
{


    public readlong(out long i)
    {
        i = long.Parse(Console.ReadLine());

    }
    public readlong(out long a, out long b)
    {
        long[] sp = Console.ReadLine().Split().Select(long.Parse).ToArray();
        a = sp[0];
        b = sp[1];

    }
    public readlong(out long a, out long b, out long c)
    {
        long[] sp = Console.ReadLine().Split().Select(long.Parse).ToArray();
        a = sp[0];
        b = sp[1];
        c = sp[2];
    }
    public readlong(out long a, out long b, out long c, out long d)
    {
        long[] sp = Console.ReadLine().Split().Select(long.Parse).ToArray();
        a = sp[0];
        b = sp[1];
        c = sp[2];
        d = sp[3];

    }

}







class Util
{




    /// <summary>
    /// 2引数の最大公約数を返します。longです。
    /// </summary>
    /// <param name="a">long 引数1</param>
    /// <param name="b">long 引数2</param>
    /// <returns></returns>
    public static long GCD(long a, long b)
    {
        if (a < b)
            swap(ref a, ref b);
        if (a % b == 0) return b;
        return GCD(b, a % b);

    }
    /// <summary>
    /// 2引数の最大公約数を返します。intです。
    /// </summary>
    /// <param name="a">int 引数1</param>
    /// <param name="b">int 引数2</param>
    /// <returns></returns>
    public static int GCD(int a, int b)
    {
        if (a < b)
            swap(ref a, ref b);
        if (a % b == 0) return b;
        return GCD(b, a % b);
    }

    /// <summary>
    /// 2つの変数の中身を交換する。型はなんでもいい
    /// </summary>
    /// <typeparam name="T"></typeparam>
    /// <param name="a"></param>
    /// <param name="b"></param>
    public static void swap<T>(ref T a, ref T b)
    {
        T temp = b;
        b = a;
        a = temp;

    }

    /// <summary>
    /// 2引数の最大公約数を返します。
    /// </summary>
    /// <param name="a"></param>
    /// <param name="b"></param>
    /// <returns></returns>
    public static long LCM(long a, long b)
    {
        return a * b / GCD(a, b);
    }
    /// <summary>
    /// 2引数の最大公約数を返します。
    /// </summary>
    /// <param name="a"></param>
    /// <param name="b"></param>
    /// <returns></returns>
    public static long LCM(int a, int b)
    {
        return Math.BigMul(a, b) / GCD(a, b);
    }


    public static int M = (int)(Math.Pow(10, 9)) + 7;
    private static int[] factorial_modM;

    /// <summary>
    /// 2引数の積 mod Mを求める。Mは10億7。
    /// 変えられるよ。
    /// </summary>
    /// <param name="a"></param>
    /// <param name="b"></param>
    /// <returns></returns>
    public static int Multiple_modM(int a, int b)
    {
        return (int)(Math.BigMul(a, b) % M);
    }
    /// <summary>
    /// n! modMを求めるための準備を行う。引数は最大値。
    /// </summary>
    /// <param name="n"></param>
    public static void factorial_modM_Prepare(int n)
    {
        factorial_modM = new int[n + 1];
        factorial_modM[0] = 1;
        for (int i = 1; i <= n; ++i)
        {
            factorial_modM[i] = Multiple_modM(factorial_modM[i - 1], i);
        }
    }

    /// <summary>
    /// n! mod M を返す。やる前にfactorial_modM_Prepare(n)を実行しといてね。
    /// </summary>
    /// <param name="n"></param>
    /// <returns></returns>
    public static int Factorial(int n)
    {
        return factorial_modM[n];
    }


    public static int Pow(int a, int m)
    {
        switch (m)
        {
            case 0:
                return 1;
            case 1:
                return a;
            default:
                int p1 = Pow(a, m / 2);
                int p2 = Multiple_modM(p1, p1);
                return ((m % 2) == 0) ? p2 : Multiple_modM(p2, a);
        }
    }
    /// <summary>
    /// a / m Mod M を求める。
    /// </summary>
    /// <param name="a"></param>
    /// <param name="b"></param>
    /// <returns></returns>
    public static int Div(int a, int b)
    {
        return Multiple_modM(a, Pow(b, M - 2));
    }
    /// <summary>
    /// nCr mod M を求める。やる前にfactorial_modM_Prepare(n)を実行しといてね。
    /// </summary>
    /// <param name="n"></param>
    /// <param name="r"></param>
    /// <returns></returns>
    public static int nCr_modM(int n, int r)
    {
        if (r < 0) return 0;

        if (n < r) { return 0; }
        if (n == r) { return 1; }
        int res = Factorial(n);
        res = Div(res, Factorial(r));
        res = Div(res, Factorial(n - r));
        return res;
    }

}
/// <summary>
/// Union-Find木を定義する。コンストラクタの指定は、頂点の番号。
/// おすすめとしては、UnionFind<T>[] で宣言してしまう。
/// </summary>
/// <typeparam name="T">int固定でいいよね。</typeparam>
class UnionFind<T>
{
    public int tree_height;
    public UnionFind<T> parent;
    public T item
    {
        get;
        private set;
    }

    public UnionFind(T item)
    {
        tree_height = 0;
        parent = this;

    }

    /// <summary>
    /// 一番上の親を探す。
    /// 探しながら繋がり方を最適化している。
    /// UnionFindの頂点そのものを返すので、==での比較が可能。
    /// </summary>
    /// <returns></returns>
    public UnionFind<T> FindAdam()
    {
        if (parent == this) return this;
        else
        {
            UnionFind<T> ParentOfParent = parent.FindAdam();
            parent = ParentOfParent;//縦長な構成をつなぎ直している
            return ParentOfParent;
        }
    }
    /// <summary>
    /// unionfind2つを一つにつなぐ。
    /// </summary>
    /// <param name="a"></param>
    /// <param name="b"></param>
    public static void Unite(UnionFind<T> a, UnionFind<T> b)
    {
        UnionFind<T> ParentOfA = a.FindAdam();
        UnionFind<T> ParentOfB = b.FindAdam();
        if (ParentOfA.tree_height < ParentOfB.tree_height)
        {
            ParentOfA.parent = ParentOfB;
            ParentOfB.tree_height = Math.Max(ParentOfA.tree_height + 1, ParentOfB.tree_height);
        }
        else
        {
            ParentOfB.parent = ParentOfA;
            ParentOfA.tree_height = Math.Max(ParentOfB.tree_height + 1, ParentOfA.tree_height);
        }

    }

}

//Treap 平衡二分木
class Treap<T> where T : IComparable
{
    private class Node
    {
        private static Random g_rand = new Random();
        private readonly int m_rank = g_rand.Next();
        private readonly T m_value;
        private Node m_lch;
        private Node m_rch;
        private int m_count;

        public Node(T value)
        {
            m_value = value;
            m_count = 1;
        }

        private static int Count(Node node)
        {
            return (node != null) ? node.m_count : 0;
        }

        private Node Update()
        {
            m_count = Count(m_lch) + Count(m_rch) + 1;
            return this;
        }

        public static Node Merge(Node a, Node b)
        {
            if (a == null) { return b; }
            if (b == null) { return a; }

            if (a.m_rank < b.m_rank)
            {
                a.m_rch = Merge(a.m_rch, b);
                return a.Update();
            }
            else
            {
                b.m_lch = Merge(a, b.m_lch);
                return b.Update();
            }
        }

        public static Tuple<Node, Node> Split(Node t, int k)
        {
            if (t == null) { return new Tuple<Node, Node>(null, null); }

            if (k <= Count(t.m_lch))
            {
                var pair = Split(t.m_lch, k);
                t.m_lch = pair.Item2;
                return new Tuple<Node, Node>(pair.Item1, t.Update());
            }
            else
            {
                var pair = Split(t.m_rch, k - Count(t.m_lch) - 1);
                t.m_rch = pair.Item1;
                return new Tuple<Node, Node>(t.Update(), pair.Item2);
            }
        }

        public int FindIndex(T value)
        {
            int L = Count(m_lch);
            if (value.CompareTo(m_value) < 0)
            {
                return (m_lch != null) ? m_lch.FindIndex(value) : 0;
            }
            else if (value.CompareTo(m_value) > 0)
            {
                return (m_rch != null) ? m_rch.FindIndex(value) + L + 1 : L + 1;
            }
            else
            {
                return L;
            }
        }

        public T this[int i]
        {
            get
            {
                int L = Count(m_lch);
                if (i < L)
                {
                    return m_lch[i];
                }
                else if (i > L)
                {
                    return m_rch[i - L - 1];
                }
                else
                {
                    return m_value;
                }
            }
        }
    }

    private Node node;

    public void Insert(T value)
    {
        if (node != null)
        {
            int k = node.FindIndex(value);
            var pair = Node.Split(node, k);
            node = Node.Merge(Node.Merge(pair.Item1, new Node(value)), pair.Item2);
        }
        else
        {
            node = new Node(value);
        }
    }

    public int FindIndex(T value)
    {
        return node.FindIndex(value);
    }

    public T this[int i]
    {
        get
        {
            return node[i];
        }
    }
}

static class Permutation<T>
{
    private static void Search(List<T[]> perms, Stack<T> stack, T[] a)
    {
        int N = a.Length;
        if (N == 0)
        {
            perms.Add(stack.Reverse().ToArray());
        }
        else
        {
            var b = new T[N - 1];
            Array.Copy(a, 1, b, 0, N - 1);
            for (int i = 0; i < a.Length; ++i)
            {
                stack.Push(a[i]);
                Search(perms, stack, b);
                if (i < b.Length) { b[i] = a[i]; }
                stack.Pop();
            }
        }
    }
    public static IEnumerable<T[]> All(IEnumerable<T> src)
    {
        var perms = new List<T[]>();
        Search(perms, new Stack<T>(), src.ToArray());
        return perms;
    }
}



namespace CompLib.Collections
{
    #region PriorityQueue
    /// <summary>
    /// 指定した型のインスタンスを最も価値が低い順に取り出すことが可能な可変サイズのコレクションを表します．
    /// </summary>
    /// <typeparam name="T">優先度付きキュー内の要素の型を指定します．</typeparam>
    /// <remarks>内部的にはバイナリヒープによって実装されています．</remarks>
    public class PriorityQueue<T>
    {
        readonly List<T> heap = new List<T>();
        readonly Comparison<T> cmp;

        /// <summary>
        /// デフォルトの比較子を使用してインスタンスを初期化します．
        /// </summary>
        /// <remarks>この操作は O(1) で実行されます．</remarks>
        public PriorityQueue() { cmp = Comparer<T>.Default.Compare; }

        /// <summary>
        /// デリゲートで表されるような比較関数を使用してインスタンスを初期化します．
        /// </summary>
        /// <param name="comparison"></param>
        /// <remarks>この操作は O(1) で実行されます．</remarks>
        public PriorityQueue(Comparison<T> comparison) { cmp = comparison; }

        /// <summary>
        /// 指定された比較子を使用してインスタンスを初期化します．
        /// </summary>
        /// <param name="comparer"></param>
        /// <remarks>この操作は O(1) で実行されます．</remarks>
        public PriorityQueue(IComparer<T> comparer) { cmp = comparer.Compare; }

        /// <summary>
        /// 優先度付きキューに要素を追加します．
        /// </summary>
        /// <param name="item">優先度付きキューに追加される要素</param>
        /// <remarks>最悪計算量 O(log N) で実行されます．</remarks>
        public void Enqueue(T item)
        {
            var pos = heap.Count;
            heap.Add(item);
            while (pos > 0)
            {
                var par = (pos - 1) / 2;
                if (cmp(heap[par], item) <= 0)
                    break;
                heap[pos] = heap[par];
                pos = par;
            }
            heap[pos] = item;

        }

        /// <summary>
        /// 優先度付きキューから最も価値が低い要素を削除し，返します．
        /// </summary>
        /// <returns>優先度付きキューから削除された要素．</returns>
        /// <remarks>最悪計算量 O(log N) で実行されます．</remarks>
        public T Dequeue()
        {
            var ret = heap[0];
            var pos = 0;
            var x = heap[heap.Count - 1];

            while (pos * 2 + 1 < heap.Count - 1)
            {
                var lch = pos * 2 + 1;
                var rch = pos * 2 + 2;
                if (rch < heap.Count - 1 && cmp(heap[rch], heap[lch]) < 0) lch = rch;
                if (cmp(heap[lch], x) >= 0)
                    break;
                heap[pos] = heap[lch];
                pos = lch;
            }
            heap[pos] = x;
            heap.RemoveAt(heap.Count - 1);
            return ret;

        }
        /// <summary>
        ///  優先度付きキューに含まれる最も価値が低い要素を削除せずに返します．
        /// </summary>
        /// <returns>優先度付きキューに含まれる最も価値が低い要素．</returns>
        /// <remarks>この操作は O(1) で実行されます．</remarks>
        public T Peek() { return heap[0]; }

        /// <summary>
        /// 優先度付きキュー内の要素の数を取得します．
        /// </summary>
        /// <returns>優先度付キュー内にある要素の数</returns>
        /// <remarks>最悪計算量 O(1) で実行されます．</remarks>
        public int Count { get { return heap.Count; } }
        /// <summary>
        /// 優先度付きキュー内に要素が存在するかどうかを O(1) で判定します．
        /// </summary>
        /// <returns>優先度付キュー内にある要素が存在するならば true，そうでなければ　false．</returns>
        /// <remarks>この操作は O(1) で実行されます．</remarks>
        public bool Any() { return heap.Count > 0; }

        /// <summary>
        /// 優先度付きキューに含まれる要素を昇順に並べて返します．
        /// </summary>
        /// <remarks>この操作は計算量 O(N log N)で実行されます．</remarks>
        public T[] Items
        {
            get
            {
                var ret = heap.ToArray();
                Array.Sort(ret, cmp);
                return ret;
            }
        }
    }
    #endregion
}

/// <summary>
/// SEGTREEは1-BASE運用でございます
/// よろしくおねがいします
/// </summary>
/// <typeparam name="T"></typeparam>
class Segtree<T>
{
    int n;//高さ
    T[] tree;//本体
    Func<T, T, T> f;//更新関数。a -> b -> ab使ったなにか
    T exnum;//初期値
    int count;
    /// <summary>
    /// !!!WARNING!!! 1-BASEです。
    /// </summary>
    /// <param name="m">最大値。200001とか入れておくとよい</param>
    /// <param name="f">２引数の関数</param>
    /// <param name="ex">葉に書いてある値の初期値。fの関数に影響を与えない数値にしておくべき。バグり散らかす。</param>
    public Segtree(int m, Func<T, T, T> f, T ex)
    {
        this.count = 0;
        this.f = f;
        this.exnum = ex;
        n = 1;
        while (n < m) n <<= 1;
        //nは mを超える2の冪数になる　m=130000 なら、 n = 131072。
        tree = new T[(n << 1) - 1];
        //treeの大きさは、2n-1
        for (int i = 0; i < tree.Length; i++) tree[i] = ex;
        //treeは全部nで初期化される
    }
    public Segtree(int m, T ini, Func<T, T, T> f, T ex) : this(m, f, ex)
    {
        this.count = 0;
        for (int i = 0; i < m; ++i) tree[i + n - 1] = ini;
        //tree[i - 1 + n]は、
        all_update();
    }

    /// <summary>
    /// j番目の値をxにする。計算し直さないので、
    /// 入れ替えた後はall_update();を使おう
    /// </summary>
    /// <param name="j">番目</param>
    /// <param name="x">更新後の値</param>
    public void assign_without_update(int j, T x)
    {
        tree[j + n - 1] = x;
    }
    /// <summary>
    /// j番目の値をxにする。計算し直すので、ちょっと計算時間がかかる
    /// </summary>
    /// <param name="j">番目</param>
    /// <param name="x">更新後の値</param>
    public void update(int j, T x)//j番目をxにする
    {
        int i = j + n - 1;
        tree[i] = x;
        while (i > 0)
        {
            i = (i - 1) >> 1;
            tree[i] = f(tree[(i << 1) + 1], tree[(i << 1) + 2]);
        }
        this.count++;
    }

    /// <summary>
    /// 木を全部計算し直す。多分NlogN。違うかも知れないから
    /// よかったら教えて
    /// </summary>
    public void all_update()
    {
        for (int i = n - 2; i >= 0; i--)
            tree[i] = f(tree[(i << 1) + 1], tree[(i << 1) + 2]);
    }
    public T look(int i) { return tree[i + n - 1]; }

    /// <summary>
    /// j番目を消す。
    /// </summary>
    /// <param name="j"></param>
    public void delete(int j)
    {
        int i = j + n - 1;
        tree[i] = exnum;
        int moved = 0;

        while (true)
        {
            tree[i + 1] = tree[i];
            T check = tree[i];
            moved++;
            if (moved + j + 2 > count)
            {
                break;
            }

        }
        all_update();
    }


    // [s, t)
    /// <summary>
    /// s番目からt番目までの結果を返す。
    /// これも当然のように1-BASE.
    /// </summary>
    /// <param name="s"></param>
    /// <param name="t"></param>
    /// <returns></returns>
    public T run(int s, int t) { return query(s, t, 0, 0, n); }
    T query(int s, int t, int k, int l, int r)
    {
        if (r <= s || t <= l) return exnum;
        if (s <= l && r <= t) return tree[k];

        return f(query(s, t, (k << 1) + 1, l, (l + r) >> 1), query(s, t, (k + 1) << 1, (l + r) >> 1, r));
    }
}