Solved Issue #548: The Tag Game (Graph BFS)

Author: Krishna200608

Problems/Graph/Day-11/sol/Krishna200608/Solution2.cpp

@@ -0,0 +1,118 @@
+/*
+Submission Link:
+https://codeforces.com/contest/813/submission/356727323
+
+
+
+Problem: The Tag Game
+Link: https://codeforces.com/contest/813/problem/C
+Author: Krishna200608
+
+Approach:
+
+    1. The game is played on a tree with Alice starting at node 1 and Bob at node x.
+    2. Alice wants to catch Bob as quickly as possible, while Bob wants to delay.
+    3. Bob’s optimal strategy is to move to a node that he can reach before Alice
+    and that is as far from the root as possible.
+    4. Once Bob reaches such a node, Alice is forced to travel down that path
+    and back, increasing the total moves.
+    5. Use two BFS traversals to calculate distances from Alice (node 1)
+    and Bob (node x) to all nodes.
+    6. For every node Bob reaches earlier than Alice, compute 2 × distance from root.
+    7. The maximum such value is the number of moves Alice makes.
+
+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; cin >> x;
+#define ill(x) ll x; cin >> x;
+#define all(x) x.begin(),x.end()
+#define in(a) for(int i = 0; i < (int)a.size(); i++) cin >> a[i];
+#define out(a) for(int i = 0; i < (int)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 adj[200005];
+int d1[200005];
+int d2[200005];
+
+void bfs(int start, int *dist, int n) {
+    for(int i = 1; i <= n; i++) dist[i] = -1;
+    dist[start] = 0;
+
+    queue<int> q;
+    q.push(start);
+
+    while(!q.empty()) {
+        int u = q.front(); q.pop();
+        for(int v : adj[u]) {
+            if(dist[v] == -1) {
+                dist[v] = dist[u] + 1;
+                q.push(v);
+            }
+        }
+    }
+}
+
+void solve() {
+
+    inn(n) inn(x)
+
+    for(int i = 1; i <= n; i++) 
+        adj[i].clear();
+
+    for(int i = 0; i < n - 1; i++) {
+        inn(u)
+        inn(v)
+        adj[u].pb(v);
+        adj[v].pb(u);
+    }
+
+    bfs(1, d1,n);
+    bfs(x, d2, n);
+
+    int ans = 0;
+    for(int i = 1; i <= n; i++) {
+        if(d2[i] < d1[i]) {
+            ans = max(ans, 2 * d1[i]);
+        }
+    }
+
+    outt(ans);
+}
+
+signed main() {
+    ios_base::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    auto begin = chrono::high_resolution_clock::now();
+
+    int t = 1;
+    while(t--) solve();
+
+    auto end = chrono::high_resolution_clock::now();
+    auto elapsed = chrono::duration_cast<chrono::nanoseconds>(end - begin);
+    cerr << "Time measured: " << elapsed.count() * 1e-6 << "ms";
+
+    return 0;
+}