As you might have guessed that if we add SCC to our final result, the whole value of that SCC will get added. Further, what we can now do is for each SCC, we find out how many other SCC's it can visit. Here we will call a DFS so that only once we add a particular node value to its corresponding DP. That would mean dp[i] = sum of dp[adjacent nodes from i] + current node value DFS will take care that you don't add a value multiple times.