There are several test cases in the input
A test case starts with two numbers N and K. (1<=N<=10000, 1<=K<=N). The cities is denoted from 1 to N. K is the capital.

The next N-1 lines each contain three numbers X, Y, D, meaning that there is a road between city-X and city-Y and the distance of the road is D. D is a positive integer which is not bigger than 1000.
Input will be ended by the end of file.

3 11 2 101 3 20

20

// Problem#: 1024// Submission#: 2973318// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <iostream>#include <stdio.h>#include <map>#include <vector>#include <cstring>using namespace std;typedef struct node {    int from;    int edge;    int to;} Node; int longest;map<int, vector<Node> > nodes;bool visited[10001];//深度优先搜索 void DFS(int k, int distance) {	bool deeper = false; //标记是否到达了叶子节点     for (int i = 0; i < nodes[k].size(); i++) {        if (visited[nodes[k].at(i).to] == false) {                 visited[nodes[k].at(i).to] = true;            deeper = true;            DFS(nodes[k].at(i).to, distance + nodes[k].at(i).edge);//此步很关键         }       }	//如果到达了叶子节点则对该路径中的权值之和与最大距离比较     if (!deeper && distance > longest) {    	longest = distance;    }}int main() {    int n, k;        while (scanf("%d %d", &n, &k) != EOF) {//刚开始超时原来是忘了加 != EOF!!!!!     	        memset(visited, false, sizeof(visited));                //使用map容器建立邻接链表         for (int i= 0; i < n-1; i++) {              int from, edge, to;            scanf("%d %d %d", &from, &to, &edge);            Node temp;            temp.from = from;            temp.edge = edge;            temp.to = to;            nodes[from].push_back(temp);                         temp.from = to;            temp.to = from;            nodes[to].push_back(temp);        }               	longest = 0;        int distance = 0;                visited[k] = true;        DFS(k, distance);        cout << longest << endl;        nodes.clear();    }    return 0;}