E - Farthest Nodes in a Tree

本文介绍了一种算法,用于解决给定带权无向树中寻找两节点间最大距离的问题。通过两次广度优先搜索(BFS),首先找到一个最远节点,再从此节点出发找到另一个最远节点,从而确定树中最远的节点对。

E - Farthest Nodes in a Tree

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
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Status
Description
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
2
4
0 1 20
1 2 30
2 3 50
5
0 2 20
2 1 10
0 3 29
0 4 50
Sample Output
Case 1: 100
Case 2: 80


#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

const int MAXN=1e6+10;
int dis[MAXN],head[MAXN];
bool vis[MAXN];
int edgenum,Tnode,ans,n;

struct Edge{
    int from;
    int to;
    int val;
    int next;
}edge[MAXN];

void init()
{
    memset(head,-1,sizeof(head));edgenum=0;
}

void addedge(int u,int v,int w)
{
    edge[edgenum].from=u;
    edge[edgenum].to=v;
    edge[edgenum].val=w;
    edge[edgenum].next=head[u];
    head[u]=edgenum++;
}

void bfs(int s)
{

    memset(dis,0,sizeof(dis));
    memset(vis,false,sizeof(vis));
    queue<int> que;
    que.push(s);vis[s]=true;dis[s]=0;//第一个入队的要提前标记 
    ans=0;

    while(!que.empty())
    {
        int now=que.front();que.pop();//取出节点 清除 
        for(int i=head[now];i!=-1;i=edge[i].next)
        {
            int go=edge[i].to;//访问与之相邻接的节点 
            if(!vis[go])//如果没访问过 访问 
            {
                if(dis[go]<dis[now]+edge[i].val)//更新距离 
                dis[go]=dis[now]+edge[i].val;
                vis[go]=true;//标记访问过 
                que.push(go);//入队 
            }
        }
    }
    for(int i=0;i<n;++i)
    {
          if(ans<dis[i])
          {
             ans=dis[i];
             Tnode=i;
          }
    }
}

int main()
{
    int kase=0,t,u,v,w,i;
    scanf("%d",&t);
    while(t--)
    {
        init();//初始化 
        scanf("%d",&n);
        for(i=1;i<n;++i)
        {
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }
        bfs(0);//从某一点开始搜索 搜到最远点 
        bfs(Tnode)//从最远点在搜索到另一最远点 
        printf("Case %d: %d\n",++kase,ans);
    }
    return 0;
}
from collections import defaultdict, deque import sys def build_graph(file_path): """从文件构建无向图""" graph = defaultdict(list) nodes = set() with open(file_path, 'r') as f: for line in f: line = line.strip() if not line or '-' not in line: continue a, b = line.split('-', 1) # 只分割第一个'-' a, b = a.strip(), b.strip() if a and b: graph[a].append(b) graph[b].append(a) nodes.add(a) nodes.add(b) return graph, nodes def break_cycles(graph, nodes): """断开图中所有环,返回生成森林""" visited = set() parent = {} forest = defaultdict(list) # 存储生成森林 for node in nodes: if node not in visited: stack = [node] visited.add(node) parent[node] = None while stack: cur = stack.pop() for neighbor in graph[cur]: # 忽略父节点 if neighbor == parent[cur]: continue # 遇到已访问节点(回边),跳过添加 if neighbor in visited: continue # 添加树边到森林 visited.add(neighbor) parent[neighbor] = cur forest[cur].append(neighbor) forest[neighbor].append(cur) stack.append(neighbor) return forest def find_tree_diameter(component, forest): """计算树的直径和路径""" def bfs(start): dist = {} pred = {} q = deque([start]) dist[start] = 0 pred[start] = None farthest_node = start while q: cur = q.popleft() for neighbor in forest[cur]: if neighbor not in dist: dist[neighbor] = dist[cur] + 1 pred[neighbor] = cur q.append(neighbor) if dist[neighbor] > dist[farthest_node]: farthest_node = neighbor return farthest_node, dist, pred # 第一次BFS找最远端点 start = next(iter(component)) u, dist_u, _ = bfs(start) # 第二次BFS找直径 v, dist_v, pred = bfs(u) # 重建路径 path = [] cur = v while cur is not None: path.append(cur) cur = pred.get(cur, None) return len(path), path # 返回节点数和路径 def find_connected_components(forest, nodes): """在生成森林中找连通分量""" visited = set() components = [] for node in nodes: if node not in visited: comp = set() stack = [node] visited.add(node) while stack: cur = stack.pop() comp.add(cur) for neighbor in forest[cur]: if neighbor not in visited: visited.add(neighbor) stack.append(neighbor) components.append(comp) return components def main(file_path): # 1. 构建原始图 graph, all_nodes = build_graph(file_path) # 2. 断开所有环得到生成森林 forest = break_cycles(graph, all_nodes) # 3. 获取连通分量(树) components = find_connected_components(forest, all_nodes) # 4. 计算每棵树的直径 max_chain_length = 0 max_chain_path = [] for comp in components: if not comp: continue length, path = find_tree_diameter(comp, forest) if length > max_chain_length: max_chain_length = length max_chain_path = path # 5. 输出结果 print(f"最长链长度(节点数): {max_chain_length}") print(f"最长链路径: {' -> '.join(max_chain_path)}") if __name__ == "__main__": if len(sys.argv) != 2: print("G:\研究生\分子对应\99999.txt") sys.exit(1) main(sys.argv[1]) 咋输入这个文件地址啊
12-12
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