lightoj 1094 - Farthest Nodes in a Tree（树的直径）

最新推荐文章于 2019-08-31 15:44:19 发布

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本文介绍了一种求解树上两点间最大距离（即树的直径）的算法，并提供了完整的C++实现代码。该算法首先从任意节点出发找到最远节点，然后从该节点出发找到树上的另一最远节点，两节点间的距离即为所求。

Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in the tree whose distance is maximum amongst all nodes.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with an integer n (2 ≤ n ≤ 30000) denoting the total number of nodes in the tree. The nodes are numbered from 0 to n-1. Each of the next n-1 lines will contain three integers u v w (0 ≤ u, v < n, u ≠ v, 1 ≤ w ≤ 10000) denoting that node u and v are connected by an edge whose weight is w. You can assume that the input will form a valid tree.

Output

For each case, print the case number and the maximum distance.

Sample Input

Output for Sample Input

0 1 20

1 2 30

2 3 50

0 2 20

2 1 10

0 3 29

0 4 50

Case 1: 100

Case 2: 80

要求树上距离最远的两点之间的距离，也就是要求树的直径，可以从任意节点先搜出一个最远端点，再从这个端点把另一个端点搜出来，求出的距离就是树的直径。

#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

struct edge
{
    int to,cost;
};
vector<edge> e[60010];
int farthest,ans;
void dfs(int x,int pre,int dis)
{
    for(int i=0;i<e[x].size();i++)
    {
        int xx = e[x][i].to;
        if(xx == pre)
            continue;
        dfs(xx,x,dis+e[x][i].cost);
    }
    if(dis > ans)
    {
        ans = dis;
        farthest = x;
    }
}
int main(void)
{
    int T,n,i,j;
    scanf("%d",&T);
    int cas = 1;
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<=n;i++)
            e[i].clear();
        for(i=0;i<n-1;i++)
        {
            int x,y;
            edge t;
            scanf("%d%d%d",&x,&y,&t.cost);
            t.to = y;
            e[x].push_back(t);
            t.to = x;
            e[y].push_back(t);
        }
        ans = 0;
        dfs(0,-1,0);
        dfs(farthest,-1,0);
        printf("Case %d: %d\n",cas++,ans);
    }

    return 0;
}