Create Or Plus Max Explanation.txt

Author: MathProgrammer

Contests/AtCoder Regular Contest 100/Explanations/Or Plus Max Explanation.txt

@@ -0,0 +1,75 @@
+The largest sum will always be between the two largest elements. 
+
+For every mask m, let us keep track of the largest and second largest element. 
+
+Instead of finding the two largest elements in O(bits set) time for every mask, 
+we can instead build it by taking the two largest elements of all it's submasks. 
+
+------
+
+We can then build a prefix maximum over this 
+
+------
+
+void update(int n, int &largest, int &second_largest, vector <int> &A)
+{
+    if(n == largest || n == second_largest)
+    {   //cout << "Return";
+        return;
+    }
+
+    if(largest == -1 || A[n] > A[largest])
+    {
+        second_largest = largest;
+        largest = n;
+    }
+    else if(second_largest == -1 || A[n] >= A[second_largest])
+    {
+        second_largest = n;
+    }
+}
+
+int main()
+{
+    int no_of_elements;
+    cin >> no_of_elements;
+
+    int max_mask = 1 << no_of_elements;
+    vector <int> A(max_mask + 1);
+    for(int i = 0; i < max_mask; i++)
+    {
+        cin >> A[i];
+    }
+
+    vector <int> largest(max_mask, -1), second_largest(max_mask, -1), sum(max_mask);
+    largest[0] = second_largest[0] = 0;
+    for(int m = 1; m < max_mask; m++)
+    {
+        //cout << "At " << m << " ";
+        update(0, largest[m], second_largest[m], A);
+        update(m, largest[m], second_largest[m], A);
+
+        for(int bit = 0; bit < no_of_elements; bit++)
+        {
+            if(is_bit_set(m, bit))
+            {
+                int submask_without_this_bit = m^(1 << bit);
+                //cout << " At old " << submask_without_this_bit << "\n";
+                update(largest[submask_without_this_bit], largest[m], second_largest[m], A);
+                update(second_largest[submask_without_this_bit], largest[m], second_largest[m], A);
+            }
+        }
+    }
+
+    for(int m = 1; m < max_mask; m++)
+    {
+        sum[m] = max(sum[m - 1], A[largest[m]] + A[second_largest[m]]);
+    }
+
+    for(int m = 1; m < max_mask; m++)
+    {
+        cout << sum[m] << "\n";
+    }
+
+    return 0;
+}