[missread the original problem]

Find any topological sort $$$T$$$ of the given DAG $$$G$$$ using Kahn's. Note that the reverse of this topological sort $$$T'$$$ is a valid topological sort for the reversed DAG $$$G'$$$.

For a node to be a "super delivery point", both of the following conditions must be satisfied:

1. When running Kahn's on $$$G$$$ to find $$$T$$$, whenever node $$$u$$$ is in the queue, there is no other node inside the queue.
2. When running Kahn's on $$$G′$$$ to find $$$T′$$$, run it in EXACTLY the reversed order as the first time, now whenever node $$$u$$$ is in the queue, there is no other node inside the queue. (just look at this code if my explanation is not understandable: 249656377)

It's easy to prove that the conjunction of these conditions is a necessary and sufficient condition.

Can we do it in this way ?

We first perform a topological sort on the given graph. Let dp1[i] represent the no of nodes, node i has contacts to. We will initially push the nodes having indegree 0 in the queue. For these nodes we will initialize dp1[i]= 0. Now, for a node which we are processing currently (say i), we will iterate over its connections (say j) and update dp[j] += dp[i]+1 iff indegree[j]>0.

Now, we will reverse the graph and perform the above operations again using another array (say dp2), where dp2[i] represent the no of nodes, node i has contacts to.

Finally, we will check for each node, if dp1[i]+dp2[i]==n-1, it means, it has connections to every node, so we will increase the count of super delivery point.

See 1062F - Upgrading Cities.

A node is a super node if the number of children + number of parents = n — 1. Now, we have to just find out the count of children and parents of each node.

int solve(int node, vector<vector<int>> &adj, vector<int> &dp){

	if(dp[node] != -1) return dp[node];

	int x = 0;
	for(int i : adj[node]){
		x += solve(i, adj, dp) + 1;
	}

	return dp[node] = x;

}

void solution()
{
	int n, m;
	cin >> n >> m;

	vector<vector<int>> adj(n);
	vector<vector<int>> directPar(n);
	for(int i = 0; i < m; i++){
		int u, v;
		cin >> u >> v;
		u--;
		v--;
		adj[u].push_back(v);
		directPar[v].push_back(u);
	}

	vector<int> dp(n, -1);
	vector<int> children(n);
	for(int i = 0; i < n; i++){
		children[i] = solve(i, adj, dp);
	}

	vector<int> par(n, 0);
	for(int i = 0; i < n; i++)dp[i] = -1;
	
	for(int i = 0; i < n; i++){	
		par[i] = solve(i, directPar, dp);
	}
	
	int res = 0;
	for(int i = 0; i < n; i++){
		if(par[i] + children[i] == n - 1){
			res++;
		}
	}

	cout << res << endl;
	
}

Order the nodes by some topological order, suppose WLOG that the order is $$$1 \rightarrow 2 \rightarrow \ldots \rightarrow n$$$ (any edge $$$i \rightarrow j$$$ satisfies $$$i < j$$$).

Compute $$$r_i$$$ as the smallest vertex reachable from vertex $$$i$$$, which is simply its edge going to the smallest vertex (if none exists, $$$r_i = \infty$$$). Symmetrically compute $$$l_i$$$ as the largest vertex that can reach $$$i$$$, which is also a neighboring edge (if none exists, $$$l_i = -\infty$$$).

Lemma: A vertex $$$v$$$ is a "super delivery point" iff $$$r_u \leq v$$$ for all $$$u < v$$$, and $$$v \leq l_u$$$ for all $$$v < u$$$.

Proof: If the condition isn't met, WLOG there exists $$$u < v$$$ such that $$$r_u > v$$$, then by definition $$$u$$$ cannot reach $$$v$$$, and because of the topological ordering, $$$v$$$ also cannot reach $$$u$$$. If the condition is met, then for every $$$u < v$$$ you will reach $$$v$$$ if you keep walking on the path $$$u \rightarrow r_u \rightarrow r_{r_u} \rightarrow \ldots \rightarrow v$$$, and symmetrically if $$$v < u$$$ then the reverse of $$$u \rightarrow l_u \rightarrow l_{l_u} \rightarrow \ldots \rightarrow v$$$ is a path. $$$\blacksquare$$$

Therefore, compute $$$l_i,r_i$$$ in $$$O(n)$$$, then compute prefix maximum of $$$r_i$$$ and suffix minimum of $$$l_i$$$, and then a vertex can be checked in $$$O(1)$$$ if it is special, so overall $$$O(n)$$$.