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Author: BrandonTang89

Other_Tasks/CF2063_C_Remove_Exactly_Two.cpp

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+/**CF2063 C
+ * 
+ * There are only 2 different cases to consider: 
+ * - If we remove two adjacent nodes
+ * - If we remove two non-adjacent nodes
+ * 
+ * If we remove adjacent nodes, the answer is the sum of the degrees of the two nodes minus 2
+ * If we remove non-adjacent nodes, the answer is the sum of their degees minus 1
+ * 
+ * We iterate over each such case separately and find the maximum answer
+ * 
+ * Time: O(n log n), Space: O(n)
+ */
+#pragma GCC optimize("Ofast")
+#pragma GCC optimize("unroll-loops")
+#include <bits/stdc++.h>
+using namespace std;
+
+typedef long long ll;
+int n;
+multiset<ll> degs;
+int main() {
+    int tc;
+    cin >> tc;
+    while (tc--) {
+        cin >> n;
+        vector<pair<int, int>> edges;
+        vector<vector<int>> adjlist(n, vector<int>());
+        degs.clear();
+
+        vector<ll> vDeg(n, 0);
+        for (int i = 0; i < n - 1; i++) {
+            int u, v;
+            cin >> u >> v;
+            u--;
+            v--;
+
+            edges.push_back({u, v});
+            vDeg[u]++;
+            vDeg[v]++;
+            adjlist[u].push_back(v);
+            adjlist[v].push_back(u);
+        }
+
+        for (int i = 0; i < n; i++) {
+            degs.insert(vDeg[i]);
+        }
+
+        ll ans = 0;
+        for (auto &[u, v] : edges) {
+            ans = max(ans, vDeg[u] + vDeg[v] - 2);
+        }
+
+        for (int i = 0; i < n; i++) {
+            degs.erase(degs.find(vDeg[i]));
+            for (auto &v : adjlist[i]) {
+                degs.erase(degs.find(vDeg[v]));
+            }
+
+            if (!degs.empty()) ans = max(ans, vDeg[i] + *degs.rbegin() - 1);
+
+            degs.insert(vDeg[i]);
+            for (auto &v : adjlist[i]) {
+                degs.insert(vDeg[v]);
+            }
+        }
+
+        cout << ans << endl;
+    }
+
+    return 0;
+}