传送门
思路：题目需要寻找 Bob 不得不改变的最小最大次数，即相当于找 1 到 n 最短路径，算是非常裸的题了。
代码实现：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9+7;
const int N = 1e5 + 10;inline void read(int &x){char t=getchar();while(!isdigit(t)) t=getchar();for(x=t^48,t=getchar();isdigit(t);t=getchar()) x=x*10+(t^48);
}int n, m, u, v, vis[N], dis[N];
queue<int> q;
vector<int> p[N];signed main()
{IOS;cin >> n >> m;for(int i = 1; i <= m; i ++){cin >> u >> v;p[u].push_back(v);p[v].push_back(u);}q.push(1);vis[1] = 1;while(q.size()){int u = q.front(); q.pop();for(int v:p[u]){if(vis[v]) continue;vis[v] = 1;dis[v] = dis[u]+1;q.push(v);}}cout << dis[n]-1 << endl;return 0;
}