sicily 1024 Magic Island

Description

There are N cities and N-1 roads in Magic-Island. You can go from one city to any other. One road only connects two cities. One day, The king of magic-island want to visit the island from the capital. No road is visited twice. Do you know the longest distance the king can go.

Input

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.

Output

One number per line for each test case, the longest distance the king can go.

Sample Input

3 1
1 2 10
1 3 20

Sample Output

20

分析：

本題要求指定起點到全部其餘點的最長路徑，須要遍歷全部頂點，但不必定是全部的邊，DFS更好些，而後直接套用DFS不斷更新長度記錄比較得出最大值便可。

代碼：

// Problem#: 1024
// Submission#: 1788079
// 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 <vector>
#include <cstring>
using namespace std;

#define MAX 10010

struct node{
    int e,d;
    node( int a, int b ){
    e = a; d = b;
    }
};

vector<node> buffer[MAX];
bool visit[MAX];
int re;

void dfs( int v, int len ){
    if( len>re ) re = len;
    visit[v] = true;
    for( int i=0 ; i<buffer[v].size() ; i++ ){
    node temp = buffer[v][i];
    if( !visit[temp.e] ){
        visit[temp.e] = true;
        len += temp.d;
        dfs(temp.e,len);
        visit[temp.e] = false;
        len -= temp.d;
    }
    }
}

int main(){
    int n,k;
    int x,y,z;
    while( cin>>n ){
    cin >> k;
    for( int i=0 ; i<n-1 ; i++ ){
        cin >> x >> y >> z;
        buffer[x].push_back(node(y,z));
        buffer[y].push_back(node(x,z));
    }
    memset(visit,0,sizeof(visit));
    re = 0;
    dfs(k,0);
    cout << re << endl;
    for( int i=1 ; i<=n ; i++ )
        buffer[i].clear();
    }
    return 0;
}