sicily 1024 邻接矩阵与深度优先搜索解题

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

 Copy sample input to clipboard

3 1
1 2 10
1 3 20

Sample Output

20

Problem Source: ZSUACM Team Member

// 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;
}