You actually don't need DP to solve this problem.

Think of it like this, Bus Stops means that you have segments in the form $$$(li, ri)$$$. In these segments you don't need to walk.

This means that if consecutive segments (let's say $$$(li, ri)$$$ & $$$(lj, rj)$$$ assuming $$$ri \lt =lj$$$ and $$$rj \gt =ri$$$) are intersecting you don't need to walk in the segment $$$(li, rj)$$$ at all.

So now this problem becomes for the distance (1, n) how many places have no segments. That is the answer. For merging segments you can sort and merge, I think it's pretty well known.

void solve() {
    cin >> n >> k;
    vpii a(k);
    rep(i, k) cin >> a[i].fi >> a[i].se;
    sor(a);
    vpii b;
    rep(i, k) {
        if(b.empty() || b.bk.se < a[i].fi)b.pb(a[i]);
        else chkmax(b.bk.se, a[i].se);
    }
    debug(b)
    ll res = 0;
    if(b.empty()) {
        print(n-1);
        return;
    }
    res += b.ft.fi - 1;
    rep(i, sz(b)-1) {
        res += b[i+1].fi-b[i].se;
    }
    res += n - b.bk.se;
    print(res);
}