Solved Issue #547: Kefa and Park (Graph DFS)

Author: Krishna200608

Problems/Graph/Day-11/sol/Krishna200608/Solution1.cpp

@@ -0,0 +1,110 @@
+/*
+Submission Link:
+https://codeforces.com/contest/580/submission/356719898
+*/
+
+/*
+Problem: Kefa and Park
+Link: https://codeforces.com/problemset/problem/580/C
+Author: Krishna200608
+
+Problem Summary:
+Kefa starts at node 1 and wants to reach any leaf node in the tree. 
+He avoids paths that contain more than m consecutive nodes with cats. 
+The task is to count how many leaf nodes are reachable without breaking this rule.
+
+Approach:
+
+    1. Treat the given graph as a tree rooted at node 1 and start DFS from it.
+    2. While traversing, keep track of consecutive nodes having cats.
+    3. Increment the count if the current node has a cat, otherwise reset it to 0.
+    4. If at any time the consecutive cat count exceeds m, prune that path.
+    5. Identify a leaf node as a node having no children other than its parent.
+    6. Count every leaf node reached through a valid path as a safe destination.
+
+
+Complexity:
+    Time: O(N)
+    Space: O(N)
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+#define MOD 1000000007
+const int M = 1e9+7;
+const double PI = acos(-1.0);
+#define INF 1e18
+#define pb push_back
+#define ppb pop_back
+#define ll long long
+#define no cout << "NO" << endl;
+#define yes cout << "YES" << endl;
+#define ff first
+#define ss second
+#define inn(x) int x; std::cin >> x;
+#define ill(x) ll x; std::cin >> x;
+#define all(x) x.begin(),x.end()
+#define in(a) for(int i = 0; i<a.size(); i++) cin>>a[i];
+#define out(a) for(int i = 0; i<a.size(); i++) cout<<a[i]<< " " ;
+typedef vector<int> vi;
+typedef vector<ll> vll;
+#define ceil_div(n, x)  ( ((n) % (x) == 0) ? ((n) / (x)) : ((n) / (x) + 1) )
+#define debug(x) cout << "x -> " << x << endl;
+#define outt(x) cout << x << endl;
+#define endl "\n"
+ 
+vi cat;
+vector<vi> adj;
+int n, m;
+int ans = 0;
+
+void dfs(int u, int p, int consec) {
+    if(cat[u]) consec++;
+    else consec = 0;
+
+    if(consec > m) return;
+
+    bool isLeaf = true;
+    for(int v : adj[u]) {
+        if(v != p) {
+            isLeaf = false;
+            dfs(v, u, consec);
+        }
+    }
+
+    if(isLeaf) ans++;
+}
+
+void solve(){
+    cin >> n >> m;
+    cat.resize(n + 1);
+    adj.resize(n + 1);
+
+    for(int i = 1; i <= n; i++) cin >> cat[i];
+
+    for(int i = 0; i < n - 1; i++) {
+        int u, v;
+        cin >> u >> v;
+        adj[u].pb(v);
+        adj[v].pb(u);
+    }
+
+    dfs(1, 0, 0);
+    outt(ans);
+}
+
+signed main(){
+ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
+auto begin = std::chrono::high_resolution_clock::now();
+    int t=1;
+    while(t--){
+        solve();
+    }
+ 
+ 
+ 
+auto end = std::chrono::high_resolution_clock::now();
+auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
+cerr << "Time measured: " << elapsed.count() * 1e-6 << "ms";
+return 0;
+}