6.树与图的广度优先遍历 图中点的层次

 

 因为这道题的边长为1，所以可以用宽搜

#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int h[N], e[N], ne[N], idx;
int d[N];
int n, m;
void add(int a, int b) {
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx;
    idx++;
}
int bfs() {
    memset(d, -1, sizeof(d)); //表示都没有遍历过
    queue<int> q;
    q.push(1);
    d[1] = 0;
    while (q.size()) {
        int t = q.front();
        q.pop();
        for (int i = h[t]; i != -1; i = ne[i]) {
            int j = e[i];
            if (d[j] == -1) {
                d[j] = d[t] + 1;
                q.push(j);
            }
        }
    }
    return d[n];
}
int main() {
    memset(h, -1, sizeof(h));
    cin >> n >> m;
    while (m--) {
        int a, b;
        cin >> a >> b;
        add(a, b);
    }
    cout << bfs() << endl;
    return 0;
}