Add DFS solution for counting paths with cats

Problems/Graph/Day-11/sol/Vishva/soln1.cpp

@@ -0,0 +1,104 @@
+#include<bits/stdc++.h>
+#include<queue>
+using namespace std;
+typedef unordered_map<int, int> umii;
+typedef unordered_map<long long, long long> umll;
+typedef unordered_map<char, long long> umci;
+typedef vector<pair<int, int>> vpi;
+typedef vector<int> vi;
+typedef long long ll;
+typedef vector<long long> vll;
+typedef unordered_map<int , bool> umib;
+#define sum(v) accumulate(v.begin(), v.end(), 0)
+#define endl '\n'
+#define f0(i, n) for(long long i = 0; i < n; i++)
+#define f1(i, n) for(long long i = 1; i < n; i++)
+#define as(v) sort(v.begin(), v.end())
+#define all(x) (x).begin(), (x).end()
+#define pb push_back
+template<class T> umll frequency(vector<T> &v) {umll freq;for(auto &x:v) freq[x]++; return freq;}
+template<class T> umci S_frequency(vector<T> &v) {umci freq;for(auto &x:v) freq[x]++; return freq;}
+template <class T> void input(vector<T> &v){for(auto &x:v)cin>>x;}
+ll power(ll x, ll y){ ll res = 1; while (y > 0){ if (y & 1) res = (ll)(res*x); y = y>>1; x = (ll)(x*x); } return res; }
+void pvll(const vector<long long> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+void pvi(const vector<int> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+
+
+// submission link :-
+//https://codeforces.com/contest/580/submission/347582553
+
+
+void dfs(ll node,ll parent, ll cats,ll m,vector<vector<ll>>&g, vll&v,ll& ans)
+{
+    if(v[node]==1)
+    {
+        cats++;
+    }
+    else
+    {
+        cats=0;
+    }
+    if(cats>m)
+    {
+        return;
+    }
+    bool isLeaf = true;
+
+    for(auto child : g[node])
+    {
+        if(child==parent)
+        {
+            continue;
+        }
+        isLeaf = false;
+        dfs(child,node,cats,m,g,v,ans);
+    }
+    if(isLeaf==true)
+    {
+        ans++;
+    }
+}
+// approach:- 
+// used dfs over here 
+//Approach :- 
+// we will do dfs traversal and keep track of the number of consecutive cats in the path
+// if the number of consecutive cats exceeds m, we stop exploring that path
+// if we reach a leaf node, we increment the answer
+
+void solve(){
+
+//-------------INPUT-------------
+    ll n; 
+    cin >> n;
+    ll m;
+    cin >> m;
+    vll v(n+1);
+    for(ll i=1;i<=n;i++)
+    {
+        cin >> v[i];
+    }
+    vector<vector<ll>>g(n+1);
+
+    for(ll i=0;i<n-1;i++)
+    {
+        ll a,b;
+        cin >> a >> b;
+        g[a].push_back(b);
+        g[b].push_back(a);
+    }
+    ll ans=0;
+    dfs(1,-1,0,m,g,v,ans);
+
+    cout << ans << endl;
+
+//-------------CODE--------------
+
+
+
+}
+
+
+int main(){
+    //int tt; cin >> tt; while(tt--)
+{solve();};
+}