题目

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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

题目大意

N个城市、(N-1)条路的岛，K为首都。从K出发，每条路只走一遍，问走过的最大距离。

思路

用变量maxDistance存所求最大距离。DFS搜索每个路径，当当前路径大于maxDistance时，当即更新。

代码

// Copyright (c) 2015 HuangJunjie@SYSU(SNO:13331087).
// sicily 1024: http://soj.sysu.edu.cn/1024
#include <cstdio>
#include <vector>

#define MAX_SIZE 10000

using namespace std;

struct Road {
  int id_;      // road id.
  int end_;     // the other end of the road
  int length_;  // the length of the road

  Road(int id, int end, int length) {
    id_ = id;
    end_ = end;
    length_ = length;
  }
};

int maxDistance;
vector<Road> Roads[MAX_SIZE + 1];   // Roads[0] is useless
bool visited[MAX_SIZE + 1];         // visited[0] is also useless


// DFS searching the roads. for the implementation is recursive, the closed list
// have to be global.
// @Param start: a city that begin the visit.
// @Param pathLength: gets a certain path's length. if the path is longer than
// the max distance we knows, update the maxDistance.
void DFS(int start, int pathLength = 0) {
  // for each road connected to start
  for (int i = 0; i < Roads[start].size(); i++) {
    // if the road is not visited
    if (!visited[Roads[start][i].id_]) {
      // visits it
      visited[Roads[start][i].id_] = true;

      // adds the road length to the path length
      pathLength += Roads[start][i].length_;
      // update the max distance
      if (pathLength > maxDistance) maxDistance = pathLength;

      // continue the path from the other end of the road
      DFS(Roads[start][i].end_, pathLength);

      // [IMPORTANT] back tracking
      pathLength -= Roads[start][i].length_;
      visited[Roads[start][i].id_] = false;
    }
  }
}

int main() {
  // the cities' number N and the capital's number K
  int N, K;

  while (scanf("%d %d", &N, &K) != EOF) {
    // initialization.
    maxDistance = 0;
    
    for (int i = 0; i <= MAX_SIZE; i++) {
      Roads[i].clear();
      visited[i] = false;
    }

    // gets the total N-1 roads.
    int start, end, length;
    for (int i = 1; i < N; i++) {
      scanf("%d %d %d", &start, &end, &length);
      // bidirenctional roads
      Roads[start].push_back(Road(i, end, length));
      Roads[end].push_back(Road(i, start, length));
    }

    // DFS the roads from the capital.
    DFS(K);

    // output the max distance visited.
    printf("%d\n", maxDistance);
  }

  return 0;
}

参考

http://www.cnblogs.com/joyeecheung/p/3505124.html