Added All Standard Graph Algos and some questions

Data Structures and Algorithms/Graph/Bipartite_graph.cpp

@@ -0,0 +1,75 @@
+#include <bits/stdc++.h>
+#define ll long long int
+#define f first
+#define s second
+#define pii pair<int,int>
+#define pic pair<int,char>
+using namespace std;
+
+// TC = O(V+E);
+// SC = O(V+E)     +       O(V)     +    O(V)
+//    for adjlist        col array      Auxilery space for recursion
+
+int bipartite_DFS(int i, int par, vector<int>adj[], vector<int> &col){
+
+  if(col[i]==-1)col[i] = 1;
+
+  for(auto x: adj[i]){
+
+         if(col[x]==-1){  // not colored yet
+
+            col[x] = 1-col[i];
+
+            if(!bipartite_DFS(x,i,adj,col))  // if one node return false then graph will not remain  bipatitite
+              return 0;
+
+         }
+         else if(col[x]==col[i]){ // x is visited now check x has different color to it's parent i or not
+            return 0;
+         }
+      }
+
+  return 1;
+
+}
+
+int main(){
+
+  int v,e;cin>>v>>e;
+
+  vector<int>adj[v+1];
+  vector<int>vis(v,-1);
+
+  for(int i=0;i<e;i++){
+    int a,b;cin>>a>>b;
+    --a,--b;
+    adj[a].push_back(b);
+    adj[b].push_back(a);
+  }
+
+  int flag = 0;
+
+  for(int i=0;i<v;i++){
+
+    if(vis[i]==-1) {// not colored yet
+
+       if(!bipartite_DFS(i,-1,adj,vis)){
+
+        flag = 1;
+        break;
+
+       }
+    }
+  }
+
+  if(flag) //no possible
+    cout<<"IMPOSSIBLE\n";
+
+  else{
+    for(auto &x: vis)
+        cout<<x+1<<" ";
+  }
+
+ return 0;
+}
+

Data Structures and Algorithms/Graph/Dense_Graph_vs_Sparse_graph.txt

@@ -0,0 +1,6 @@
+Dense graph is a graph in which the number of edges is close to the maximal number of edges. 
+Sparse graph is a graph in which the number of edges is close to the minimal number of edges. Sparse graph can be a disconnected graph.
+
+As the names indicate sparse graphs are sparsely connected (eg: Trees). Usually the number of edges is in O(n) where n is the number of vertices. Therefore adjacency lists are preferred since they require constant space for every edge.
+
+Dense graphs are densely connected. Here number of edges is usually O(n^2). Therefore adjacency matrix is preferred.

Data Structures and Algorithms/Graph/Maximum_Diameter_in_N_ary_tree.cpp.cpp

@@ -0,0 +1,58 @@
+#include<bits/stdc++.h>
+using namespace std;
+#define FASTIO  ios_base::sync_with_stdio(false);cin.tie(NULL);
+#define f first
+#define s second
+#define pb push_back 
+#define ll long long
+int MOD = 1e9+7;
+
+int ans=0;
+int DFS(int i,int par,vector<int>adj[]){
+
+    int mx  = 0, premax = 0;
+    for(auto x: adj[i]){
+        if(x != par){
+            int height = DFS(x,i,adj);
+            if(mx < height )
+                premax = mx, mx = height;
+            else if(premax < height)
+                premax = height;
+        }
+    }
+     ans = max(ans, mx+premax+1);
+    return 1 + mx;
+}
+int main(){
+
+    int n;cin>>n; // number of nodes
+    vector<int>adj[n+1];
+
+    for(int i=0;i<n-1;i++){
+       int u, v; cin>>v>>u;
+       --v,--u;
+       adj[v].emplace_back(u);
+       adj[u].emplace_back(v);
+    }
+
+    DFS(0,-1,adj);
+    cout<<ans;
+}
+
+/*
+
+Input
+11         // numbr of vertices
+1 2        // next n-1 lines of edges connected u to v
+1 7
+1 6 
+1 4 
+1 5
+2 10
+10 11
+2 3
+7 8 
+8 9
+
+output 
+7

Data Structures and Algorithms/Graph/Robber_&_you.cpp

@@ -0,0 +1,59 @@
+#include<bits/stdc++.h>
+using namespace std;
+#define pb push_back 
+#define ll long long
+
+vector<int>gr[100002];
+int vis[100002],par[100002],c[100002];
+
+void dfs(int a,int p){
+    par[a]=p;
+    for(int i=0;i<gr[a].size();i++)
+        if(gr[a][i]!=p)
+            dfs(gr[a][i],a);
+}
+
+int dfs1(int a,int p){
+    int ans =- 100000;
+    for(int i=0;i<gr[a].size();i++)
+        if(gr[a][i]!=p)
+            ans=max(ans,dfs1(gr[a][i],a));
+    if(ans==-100000)ans=0;
+    return ans+c[a];
+}
+
+int main()
+{
+    int n,r,u,v;
+    cin>>n>>r;
+    for(int i=1;i<n;i++){
+        cin>>u>>v;
+        gr[u].pb(v);
+        gr[v].pb(u);
+    }
+    // int c[n+1];
+    for(int i=1;i<=n;i++){
+        cin>>c[i];
+        vis[i]=0;
+    }
+    dfs(1,1);
+    vector<int>vv;   // storing the path
+
+    while(r!=1){
+        vv.pb(r);
+        r=par[r];
+    }
+
+    vv.pb(1);
+
+   for(auto x: vv)cout<<x<<" ";
+
+    for(int i=0;i<(vv.size())/2;i++)
+        c[vv[i]]=0;                  // half of the path will be of robber
+    if((vv.size()) % 2 == 1)         // if total length of the path is odd then they will definately meet pathLen/2  
+        c[vv[(vv.size())/2]]/=2;
+    cout<<dfs1(1,1);
+
+    return 0;
+
+}

Data Structures and Algorithms/Graph/Rotten_Oranges_(Multisource_BFS).cpp

@@ -0,0 +1,67 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// start BFS from all rotten oranges initiallly and Cnt time in every level
+// until all oranges becomes rottten
+int inf  = 1e5;
+
+int orangesRotting(vector<vector<int>>& grid) {
+
+    int r = grid.size(), c = grid[0].size();
+
+    queue<pair<int,int>>q;
+    vector<vector<int>>dis(r+1,vector<int>(c+1,inf));
+    int time=0;
+
+    // Multisource BFS
+    for(int i=0;i<r;i++){
+        for(int j=0;j<c;j++){
+            if(grid[i][j]==2){
+                q.push({i,j});  //pushing all the rotten oranges intially in the queue
+                dis[i][j] = 0;  // make 0 for all the starting points
+            }
+        }
+    }
+
+    int dx[] = {0,0,-1,1};
+    int dy[] = {1,-1,0,0};
+
+    while(!q.empty()){  
+
+        int i = q.front().first, j = q.front().second;
+        q.pop();
+
+        for(int k=0;k<4;k++){
+            int nx = i + dx[k], ny = j + dy[k];
+            // going for fresh oranges to all its adjacent 
+            if(nx>=0&&nx<r&&ny>=0&&ny<c && grid[nx][ny]==1 && dis[nx][ny]==inf){
+
+                // if  we get fresh orange then mark it rotten and push into queue
+                // update the distance of (nx,ny) means we can reach (nx,ny) from                               (i,j) in one step;     
+
+                    q.push({nx,ny}); 
+                    dis[nx][ny] = dis[i][j]+1;
+                    time = max(time, dis[nx][ny]);
+            }     
+        }      
+    }
+
+    // If any rotten orange is remaining where we can't reach return -1
+     for(int i=0;i<r;i++){
+        for(int j=0;j<c;j++){
+            if(grid[i][j]==1 && dis[i][j]==inf){  
+                return -1;
+            }
+        }
+    } 
+    return time;
+}
+
+
+
+// Input: 
+// grid = [[2,1,1],
+//         [1,1,0],
+//         [0,1,1]]
+
+// Output: 4