CF868E.Policeman and a Tree

普及/提高-

通过率:0%

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题目描述

You are given a tree (a connected non-oriented graph without cycles) with vertices numbered from 11 to nn , and the length of the ii -th edge is wiw_{i} . In the vertex ss there is a policeman, in the vertices x1,x2,...,xmx_{1},x_{2},...,x_{m} ( xjsx_{j}≠s ) mm criminals are located.

The policeman can walk along the edges with speed 11 , the criminals can move with arbitrary large speed. If a criminal at some moment is at the same point as the policeman, he instantly gets caught by the policeman. Determine the time needed for the policeman to catch all criminals, assuming everybody behaves optimally (i.e. the criminals maximize that time, the policeman minimizes that time). Everybody knows positions of everybody else at any moment of time.

输入格式

The first line contains single integer nn ( 1<=n<=501<=n<=50 ) — the number of vertices in the tree. The next n1n-1 lines contain three integers each: uiu_{i} , viv_{i} , wiw_{i} ( 1<=ui,vi<=n1<=u_{i},v_{i}<=n , 1<=wi<=501<=w_{i}<=50 ) denoting edges and their lengths. It is guaranteed that the given graph is a tree.

The next line contains single integer ss ( 1<=s<=n1<=s<=n ) — the number of vertex where the policeman starts.

The next line contains single integer mm ( 1<=m<=501<=m<=50 ) — the number of criminals. The next line contains mm integers x1,x2,...,xmx_{1},x_{2},...,x_{m} ( 1<=xj<=n1<=x_{j}<=n , xjsx_{j}≠s ) — the number of vertices where the criminals are located. xjx_{j} are not necessarily distinct.

输出格式

If the policeman can't catch criminals, print single line "Terrorists win" (without quotes).

Otherwise, print single integer — the time needed to catch all criminals.

输入输出样例

  • 输入#1

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

    输出#1

    8
    
  • 输入#2

    6
    1 2 3
    2 3 5
    3 4 1
    3 5 4
    2 6 3
    2
    3
    1 3 5
    

    输出#2

    21
    

说明/提示

In the first example one of the optimal scenarios is the following. The criminal number 22 moves to vertex 33 , the criminal 44 — to vertex 44 . The policeman goes to vertex 44 and catches two criminals. After that the criminal number 11 moves to the vertex 22 . The policeman goes to vertex 33 and catches criminal 22 , then goes to the vertex 22 and catches the remaining criminal.

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