为什么在 BFS 末尾打印 1？

Question

下面是无向图上的简单广度优先搜索。问题是找出源节点和目标节点之间是否存在路径。代码有效，但我不明白为什么要在最后打印 1。

节目：

#include <iostream>
#include <unordered_map>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

bool BFS(unordered_map<int, vector<int>> &umap, int source, int dest) {
    
    queue<int> q;
    vector<int> v;
    q.push(source);
    
    while (!q.empty()) {
        int front = q.front();

        if (find(v.begin(), v.end(), front) == v.end()) {//element is not visited yet
            cout << static_cast<char>(front) << " -> ";
            if (front == dest)
                return true;
            v.push_back(front);

            for (auto &i: umap[front])
                q.push(i);
        }
        q.pop();
    }
    return false;
}
int main() {
    unordered_map<int, vector<int>> umap;
    umap['i'] = {'j', 'k'}; 
    umap['j'] = {'i'};
    umap['k'] = {'i', 'm', 'p'};
    umap['m'] = {'k'};
    umap['p'] = {'k'};
    umap['o'] = {'n'};
    umap['n'] = {'o'};
    
    cout << endl
         << "------------BFS------------"
         << endl;
    cout << BFS(umap, 'j', 'm');
    
}

输出：

------------BFS------------
j -> i -> k -> m -> 1
Process finished with exit code 0

Answer 1

关于这个问题的第一条评论总结了这一点——我忽略了我正在打印 bool (cout << BFS(umap, 'j', 'm');) 的事实，所以在这种情况下预期会出现 1找到路径（值为true）。

为什么在 BFS 末尾打印 1？

Why is a 1 being printed at the end of BFS?

c++

breadth-first-search