Lightoj 1094 - Farthest Nodes in a Tree 【树的直径 裸题】

本文详细阐述了如何计算树的直径,即寻找树中距离最远的两个节点,并提供了AC代码实现。此过程涉及图论基础知识,适用于计算机科学领域的算法学习与实践。

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1094 - Farthest Nodes in a Tree
Time Limit: 2 second(s)Memory Limit: 32 MB

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-1lines 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

2

4

0 1 20

1 2 30

2 3 50

5

0 2 20

2 1 10

0 3 29

0 4 50

Case 1: 100

Case 2: 80

Notes

Dataset is huge, use faster i/o methods.



题意:求树的直径,水题。。。



AC代码:


#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define MAXN 30000+10
#define MAXM 60000+10
#define INF 1000000
#define eps 1e-8
using namespace std;
struct Edge
{
    int from, to, val, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int dist[MAXN];
bool vis[MAXN];
int node, ans;
int N;
int k = 1;
void init()
{
    edgenum = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w)
{
    Edge E = {u, v, w, head[u]};
    edge[edgenum] = E;
    head[u] = edgenum++;
}
void getMap()
{
    int a, b, c;
    for(int i = 1; i < N; i++)
    {
        scanf("%d%d%d", &a, &b, &c);
        a++, b++;
        addEdge(a, b, c);
        addEdge(b, a, c);
    }
}
void BFS(int sx)
{
    queue<int> Q;
    memset(dist, 0, sizeof(dist));
    memset(vis, false, sizeof(vis));
    vis[sx] = true;
    ans = 0;
    node = sx;
    Q.push(sx);
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            Edge E = edge[i];
            if(!vis[E.to] && dist[E.to] < dist[u] + E.val)
            {
                vis[E.to] = true;
                dist[E.to] = dist[u] + E.val;
                if(dist[E.to] > ans)
                {
                    ans = dist[E.to];
                    node = E.to;
                }
                Q.push(E.to);
            }
        }
    }
}
void solve()
{
    BFS(1);
    BFS(node);
    printf("Case %d: ", k++);
    printf("%d\n", ans);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        init();
        getMap();
        solve();
    }
    return 0;
}



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