LightOJ 1094-1094 - Farthest Nodes in a Tree【树的直径模板】

墨蓝 2022-08-21 09:42 83阅读 0赞

1094 - Farthest Nodes in a Tree

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   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Time Limit:&nbsp;<span style="color:rgb(180,95,4)">2 second(s)</span></td> 
   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Memory Limit:&nbsp;<span style="color:rgb(180,95,4)">32 MB</span></td> 
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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.

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   <td> <h1 style="color:green">Sample Input</h1> </td> 
   <td> <h1 style="color:green">Output for Sample Input</h1> </td> 
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   <td> <p>2</p> <p>4</p> <p>0 1 20</p> <p>1 2 30</p> <p>2 3 50</p> <p>5</p> <p>0 2 20</p> <p>2 1 10</p> <p>0 3 29</p> <p>0 4 50</p> </td> 
   <td> <p>Case 1: 100</p> <p>Case 2: 80</p> </td> 
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# Notes #

Dataset is huge, use faster i/o methods.

解题思路：

直接套模板，数组要开大，要用邻接表。

#include <iostream>
    #include<stdio.h>
    #include <string.h>
    #define MAXN 60005
    #define INF 0x3f3f3f3f
    
    using namespace std;
    
    struct Edge
    {
        int v, w, next;
    }edge[MAXN*6];
    
    int head[MAXN], vis[MAXN], d[MAXN], q[MAXN], e, n;
    
    int init()
    {
        e = 0;
        memset(head, -1, sizeof(head));
    }
    
    void add(int u, int v, int w)
    {
        edge[e].v = v;
        edge[e].w = w;
        edge[e].next = head[u];
        head[u] = e++;
    }
    
    void bfs(int src)
    {
        for (int i = 0; i <= n; i++)
        {
            vis[i] = 0;
            d[i] = INF;
        }
    
        int h = 0, t = 0;
        vis[src] = 1;
        q[t++] = src;
        d[src] = 0;
        while (h < t)
        {
            int u = q[h++];
            for (int i = head[u]; i != -1; i = edge[i].next)
            {
                int v = edge[i].v;
                int w = edge[i].w;
                if (d[u] + w < d[v])
                {
                    d[v] = d[u] + w;
                    if (!vis[v])
                    {
                        q[t++] = v;
                        vis[v] = 1;
                    }
                }
            }
        }
    }
    
    int main()
    {
        int u, v, w;
        int t;
    	scanf("%d",&t);
    	int cnm=1;
    	while(t--)
    	{
    		init();
    	    scanf("%d",&n);
    	    for (int i = 1; i <= n - 1; i++)
    	    {
    	        scanf("%d%d%d",&u,&v,&w);
    	        add(u, v, w);
    	        add(v, u, w);
    	    }
    	    bfs(1);
    	    int pos = 0, mx = -1;
    	    for (int i  = 0; i < n; i++)
    	    {
    	        if(d[i] > mx)
    	        {
    	            mx = d[i];
    	            pos = i;
    	        }
    	    }
    	    bfs(pos);
    	    mx = -1;
    	    for (int i  = 0; i < n; i++)
    	    {
    	        if(d[i] > mx)
    	        {
    	            mx = d[i];
    	        }
    	    }
    	    printf("Case %d: ",cnm++);
    	    printf("%d\n",mx);
    	} 
        return 0;
    }