Added more stuff

Author: Mehulcoder

Advanced Graphs/Bottom.cpp

@@ -80,6 +80,7 @@ void dfs(vector<int>* edges, int start, int* visited, stack<int> &s){
 }
 
 unordered_set<unordered_set<int>*>* kosaraju(vector<int>* edges, vector<int>* edgest, int v){
+	
 	//Initialize an empty set of sets for the answer
 	unordered_set<unordered_set<int>*>* ans = new unordered_set<unordered_set<int>*>();
 
@@ -137,6 +138,7 @@ int main( int argc , char ** argv )
 	cin.tie(NULL) ;
 
 	while(1){
+
 		int v, e;
 		cin>>v;
 		if (v==0)
@@ -157,31 +159,66 @@ int main( int argc , char ** argv )
 		}
 
 		unordered_set<unordered_set<int>*>* components = kosaraju(edges, edgest, v);
+		// auto itt = components->begin();
+		// while(itt!=components->end()){
+		// 	auto itt2 = (*itt)->begin();
+		// 	while(itt2!=(*itt)->end()){
+		// 		cout << *itt2+1 << ' ';
+		// 		itt2++;
+		// 	}
+		// 	itt++;
+		// 	cout << '\n';
+		// }
+		
+	//	cout << (*components).size() << '\n';		
 
 		auto it = components->begin();
-		while(it!=components->end()){
+		vector<int> ans;
+		while(it!=components->end())
+		{	
+			int flag = 0;
 			auto it2 = (*it)->begin();
-			while(it2!=(*it)->end()){
-				int i = 0;
-				for (i = 0; i < edges[*it2].size(); ++i)
+			while(it2!=(*it)->end())
+			{		
+				//int i = 0;
+				for (int i = 0; i < edges[*it2].size(); ++i)
 				{
-					if (it->count(edges[*it2].at(i)) == 0)
+
+					if ((*it)->count(edges[*it2].at(i)) == 0)
 					{
+						flag = 1;
 						break;
+
 					}
 				}
-				if (i==edges[*it2].size())
+				if (flag == 1)
 				{
 					break;
 				}
 				it2++;
 			}
-			while(it2!=(*it)->end()){
-				cout << *it2 << ' ';
+			if (flag == 0)
+			{
+				//vector<int> ans;
+				it2 = (*it)->begin();
+				while(it2!=(*it)->end())
+				{
+					ans.push_back(*it2 + 1);
+					//cout << *it2 + 1<< ' ';
+					it2++;
+				}
+				
+				//cout <<'\n';
+				//break;
 			}
-			cout <<'\n';
+			
 			it++;
 		}
+		sort(ans.begin(), ans.end());
+		for (int i = 0; i < ans.size(); ++i)
+		{
+			cout<<ans.at(i)<<" ";
+		}
 	}
 
 

Advanced Graphs/Monk.cpp

@@ -0,0 +1,83 @@
+/*
+
+				Name: Mehul Chaturvedi
+				IIT-Guwahati
+
+*/
+
+
+/*
+MONK AND THE ISLAND
+Monk visits the land of Islands. There are a total of N islands numbered from 1 to N. Some pairs of islands are connected to each other by Bidirectional bridges running over water.
+Monk hates to cross these bridges as they require a lot of efforts. He is standing at Island #1 and wants to reach the Island #N. Find the minimum the number of bridges that he shaint have to cross, if he takes the optimal route.
+Input:
+First line contains T. T testcases fointow.
+First line of each test case contains two space-separated integers N, M.
+Each of the next M lines contains two space-separated integers X and Y , denoting that there is a bridge between Island X and Island Y.
+Output:
+Print the answer to each test case in a new line.
+Constraints:
+1 ≤ T ≤ 10
+1 ≤ N ≤ 10000
+1 ≤ M ≤ 100000
+1 ≤ X, Y ≤ N
+SAMPLE INPUT
+2
+3 2
+1 2
+2 3
+4 4
+1 2
+2 3
+3 4
+4 2
+SAMPLE OUTPUT
+2
+2
+*/
+
+using namespace std;
+#define pencho ios_base::sync_with_stdio(0);cin.tie(0); 
+#define mod 1000000007
+int n;
+int dp[10001];
+int solve(vector<int>adj[])
+{
+    queue<int>q;
+    dp[1]=0;
+    q.push(1);
+    while(!q.empty())
+    {
+        int idx=q.front();
+        q.pop();
+        for(int i=0;i<adj[idx].size();i++)
+        {
+            if(dp[adj[idx][i]]==-1)
+            {
+                dp[adj[idx][i]]=dp[idx]+1;
+                q.push(adj[idx][i]);
+            }
+        }
+    }
+    return dp[n];
+}
+int main()
+{
+    pencho
+    int t,m;
+    cin>>t;
+    while(t--)
+    {
+        cin>>n>>m;
+        memset(dp,-1,sizeof(dp));
+        vector<int>adj[n+1];
+        while(m--)
+        {
+            int x,y;
+            cin>>x>>y;
+            adj[x].push_back(y);
+            adj[y].push_back(x);
+        }
+        cout<<solve(adj)<<endl;
+    }
+}

Advanced Graphs/Test.txt

@@ -1,3 +1,7 @@
-3 3
-1 3 2 3 3 1
-0
+1
+4 5
+1 2
+1 3
+2 3
+2 4
+3 4