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这种广度优先搜索的实现有什么问题?

[英]What's wrong with this implementation of Breadth First Search?

Here is my whole program:这是我的整个程序:

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

void addEdge(vector<int> adjList[], int u, int v) {
    adjList[u].push_back(v);
    adjList[v].push_back(u);
}

void bfs(int s, vector<int> adjList[], vector<bool> visited, int V) {
    queue<int> q;
    q.push(s);
    visited[s] = true;
    while (!q.empty()) {
        int cur = q.front();
        q.pop();

        cout << cur << " ";
        for (int i = 0; i < adjList[cur].size(); i++) {
            if (!visited[i]) {
                q.push(adjList[cur][i]);
                visited[i] = true;
            }
        }
    }
}

int main() {
    int V = 4;
    vector<int> adj[4];
    vector<bool> visited;
    visited.assign(V, false);
    addEdge(adj, 0, 1);
    addEdge(adj, 2, 3);
    addEdge(adj, 3, 4);
    bfs(0, adj, visited, V);
    return 0;
}

I want to perform Breadth First Search (BFS) and print out nodes as they are discovered.我想执行广度优先搜索 (BFS) 并在发现节点时打印出它们。 For some reason this isn't printing anything.出于某种原因,这不会打印任何内容。 What have I done wrong?我做错了什么?

  1. You have out-of-bounds access to adj .您可以越界访问adj It has size 4 , but in addEdge(adj, 3, 4);它的大小为4 ,但在addEdge(adj, 3, 4); you're writing into the 4-th element (5-th in one-based indexing).您正在写入第 4 个元素(基于一个的索引中的第 5 个)。

  2. The condition if (!visited[i]) doesn't make sense.条件if (!visited[i])没有意义。 i should be replaced with the next node index, ie adjList[cur][i] . i应该替换为下一个节点索引,即adjList[cur][i]

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