|
| 1 | +The largest sum will always be between the two largest elements. |
| 2 | + |
| 3 | +For every mask m, let us keep track of the largest and second largest element. |
| 4 | + |
| 5 | +Instead of finding the two largest elements in O(bits set) time for every mask, |
| 6 | +we can instead build it by taking the two largest elements of all it's submasks. |
| 7 | + |
| 8 | +------ |
| 9 | + |
| 10 | +We can then build a prefix maximum over this |
| 11 | + |
| 12 | +------ |
| 13 | + |
| 14 | +void update(int n, int &largest, int &second_largest, vector <int> &A) |
| 15 | +{ |
| 16 | + if(n == largest || n == second_largest) |
| 17 | + { //cout << "Return"; |
| 18 | + return; |
| 19 | + } |
| 20 | + |
| 21 | + if(largest == -1 || A[n] > A[largest]) |
| 22 | + { |
| 23 | + second_largest = largest; |
| 24 | + largest = n; |
| 25 | + } |
| 26 | + else if(second_largest == -1 || A[n] >= A[second_largest]) |
| 27 | + { |
| 28 | + second_largest = n; |
| 29 | + } |
| 30 | +} |
| 31 | + |
| 32 | +int main() |
| 33 | +{ |
| 34 | + int no_of_elements; |
| 35 | + cin >> no_of_elements; |
| 36 | + |
| 37 | + int max_mask = 1 << no_of_elements; |
| 38 | + vector <int> A(max_mask + 1); |
| 39 | + for(int i = 0; i < max_mask; i++) |
| 40 | + { |
| 41 | + cin >> A[i]; |
| 42 | + } |
| 43 | + |
| 44 | + vector <int> largest(max_mask, -1), second_largest(max_mask, -1), sum(max_mask); |
| 45 | + largest[0] = second_largest[0] = 0; |
| 46 | + for(int m = 1; m < max_mask; m++) |
| 47 | + { |
| 48 | + //cout << "At " << m << " "; |
| 49 | + update(0, largest[m], second_largest[m], A); |
| 50 | + update(m, largest[m], second_largest[m], A); |
| 51 | + |
| 52 | + for(int bit = 0; bit < no_of_elements; bit++) |
| 53 | + { |
| 54 | + if(is_bit_set(m, bit)) |
| 55 | + { |
| 56 | + int submask_without_this_bit = m^(1 << bit); |
| 57 | + //cout << " At old " << submask_without_this_bit << "\n"; |
| 58 | + update(largest[submask_without_this_bit], largest[m], second_largest[m], A); |
| 59 | + update(second_largest[submask_without_this_bit], largest[m], second_largest[m], A); |
| 60 | + } |
| 61 | + } |
| 62 | + } |
| 63 | + |
| 64 | + for(int m = 1; m < max_mask; m++) |
| 65 | + { |
| 66 | + sum[m] = max(sum[m - 1], A[largest[m]] + A[second_largest[m]]); |
| 67 | + } |
| 68 | + |
| 69 | + for(int m = 1; m < max_mask; m++) |
| 70 | + { |
| 71 | + cout << sum[m] << "\n"; |
| 72 | + } |
| 73 | + |
| 74 | + return 0; |
| 75 | +} |
0 commit comments