Sum of dependencies in a graph - BASIC.cpp

/*Sum of dependencies in a graph 

Basic Accuracy: 79.22% Submissions: 1246 Points: 1
Given a directed graph with V nodes and E edges. If there is an edge from u to v then u depends on v. Find out the sum of dependencies for every node. Duplicate edges should be counted as separate edges.

Example 1:

Input:
V=4
E=4
Edges={ {0,2},{0,3},{1,3},{2,3} }

Output:
4
Explanation:
For the graph in diagram, A depends
on C and D i.e. 2, B depends on D i.e.
1, C depends on D i.e. 1
and D depends on none.
Hence answer -> 0 + 1 + 1 + 2 = 4
Example 2:

Input:
V=4
E=3
Edges={ {0,3},{0,2},{0,1} }
Output:
3
Explanation:
The sum of dependencies=3+0+0+0=3.

Your Task:
You don't need to read input or print anything.Your task is to complete the function sumOfDependencies() which takes the adj (Adjacency list) and 
V (Number of nodes)as input parameters and returns the total sum of dependencies of all nodes.


Expected Time Complexity:O(V)
Expected Auxillary Space:O(1)


Constraints:
1<=V,E<=150

0<= Edges[i][0],Edges[i][1] <= V-1
*/


//DRIVER CODE
#include<bits/stdc++.h>
using namespace std;
vector<vector<int>> g;
void addedge(int a,int b){
    g[a].push_back(b);
}

//YOUR CODE
int find(int n){
    int sum=0;
    for(auto i=0;i<n;i++)
     sum+=g[i].size();
    return sum;
}


//DRIVER CODE
int main() {
	//code
	int t,n,m,a,b;
	cin>>t;
	while(t--){
	    cin>>n>>m;
	    g.assign(n,vector<int>());
	    for(int i=0;i<m;i++)
	    {
	        cin>>a>>b;
	        addedge(a,b);
	    }
	    cout<<find(n)<<endl;
	}
	return 0;
}