Created
October 31, 2020 22:36
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Check if Graph is Forest or Not [TC: O(V+E); SC: O(V)]
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// Forest Definition: Disjoint Union of Acyclic Graphs (or Trees) | |
#include <vector> | |
#include <iostream> | |
#include <algorithm> | |
using namespace std; | |
bool DFSRec(vector<int> *G, vector<bool> &visited, int curr, int parent) { | |
visited[curr] = true; | |
for(int adj: G[curr]) { | |
if(!visited[adj]) { | |
if(DFSRec(G, visited, adj, curr)) | |
return true; | |
} | |
else if(adj != parent) | |
return true; | |
} | |
return false; | |
} | |
bool DFS(vector<int> *G, const int V) { | |
vector<bool> visited(V, false); | |
for(int i = 0; i < V; ++i) | |
if(!visited[i]) | |
if(DFSRec(G, visited, i, -1)) | |
return true; | |
return false; | |
} | |
void addEdge(vector<int> *G, int u, int v) { | |
G[u].push_back(v); | |
G[v].push_back(u); | |
} | |
int main() { | |
int t; cin >> t; | |
while(t--) { | |
int V, E, u, v; cin >> V >> E; | |
vector<int> G[V]; | |
for(int i = 0; i < E; ++i) { | |
cin >> u >> v; | |
addEdge(G, u-1, v-1); | |
} | |
cout << (DFS(G, V) ? "No" : "Yes") << "\n"; | |
} | |
return 0; | |
} | |
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