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| 1 | +// Time: O(C(m * n, i) * C(m * n - i, e)) |
| 2 | +// Space: O(min(m * n, i + e)) |
| 3 | + |
| 4 | +class Solution { |
| 5 | +public: |
| 6 | + int getMaxGridHappiness(int m, int n, int introvertsCount, int extrovertsCount) { |
| 7 | + return iter_backtracking(m, n, introvertsCount, extrovertsCount); |
| 8 | + } |
| 9 | + |
| 10 | +private: |
| 11 | + int iter_backtracking(int m, int n, int i, int e) { |
| 12 | + int result = 0; |
| 13 | + vector<int> curr; |
| 14 | + vector<tuple<int, int, int, int, int>> stk = {{2, i, e, 0, 0}}; |
| 15 | + while (!empty(stk)) { |
| 16 | + const auto [step, i, e, total, x] = stk.back(); stk.pop_back(); |
| 17 | + if (step == 2) { |
| 18 | + if ((size(curr) == m * n) || (i == 0 && e == 0)) { |
| 19 | + result = max(result, total); |
| 20 | + continue; |
| 21 | + } |
| 22 | + if (total + (i + e) * 120 < result) { // pruning |
| 23 | + continue; |
| 24 | + } |
| 25 | + if (e > 0) { |
| 26 | + int new_total = count_total(n, curr, 2, total); |
| 27 | + stk.emplace_back(3, 0, 0, 0, 0); |
| 28 | + stk.emplace_back(2, i, e - 1, new_total, 0); |
| 29 | + stk.emplace_back(1, 0, 0, 0, 2); |
| 30 | + } |
| 31 | + if (i > 0) { |
| 32 | + int new_total = count_total(n, curr, 1, total); |
| 33 | + stk.emplace_back(3, 0, 0, 0, 0); |
| 34 | + stk.emplace_back(2, i - 1, e, new_total, 0); |
| 35 | + stk.emplace_back(1, 0, 0, 0, 1); |
| 36 | + } |
| 37 | + if (left(n, curr) || up(n, curr)) { // leave unoccupied iff left or up is occupied |
| 38 | + stk.emplace_back(3, 0, 0, 0, 0); |
| 39 | + stk.emplace_back(2, i, e, total, 0); |
| 40 | + stk.emplace_back(1, 0, 0, 0, 0); |
| 41 | + } |
| 42 | + } else if (step == 1) { |
| 43 | + curr.emplace_back(x); |
| 44 | + } else if (step == 3) { |
| 45 | + curr.pop_back(); |
| 46 | + } |
| 47 | + } |
| 48 | + return result; |
| 49 | + } |
| 50 | + |
| 51 | + int left(int n, const vector<int>& curr) { |
| 52 | + return (size(curr) % n) ? curr[size(curr) - 1] : 0; |
| 53 | + } |
| 54 | + |
| 55 | + int up(int n, const vector<int>& curr) { |
| 56 | + return (size(curr) >= n) ? curr[size(curr) - n] : 0; |
| 57 | + } |
| 58 | + |
| 59 | + int count_total(int n, const vector<int>& curr, int t, int total) { |
| 60 | + return (total |
| 61 | + - 30 * ((left(n, curr) == 1) + (up(n, curr) == 1)) |
| 62 | + + 20 * ((left(n, curr) == 2) + (up(n, curr) == 2)) |
| 63 | + + (120 - 30 * ((left(n, curr) != 0) + (up(n, curr) != 0))) * (t == 1) |
| 64 | + + ( 40 + 20 * ((left(n, curr) != 0) + (up(n, curr) != 0))) * (t == 2)); |
| 65 | + } |
| 66 | +}; |
| 67 | + |
| 68 | +// Time: O(C(m * n, i) * C(m * n - i, e)) |
| 69 | +// Space: O(min(m * n, i + e)) |
| 70 | +class Solution2 { |
| 71 | +public: |
| 72 | + int getMaxGridHappiness(int m, int n, int introvertsCount, int extrovertsCount) { |
| 73 | + int result = 0; |
| 74 | + vector<int> curr; |
| 75 | + backtracking(m, n, introvertsCount, extrovertsCount, 0, &curr, &result); |
| 76 | + return result; |
| 77 | + } |
| 78 | + |
| 79 | +private: |
| 80 | + void backtracking(int m, int n, int i, int e, int total, |
| 81 | + vector<int> *curr, int *result) { |
| 82 | + if ((size(*curr) == m * n) || (i == 0 && e == 0)) { |
| 83 | + *result = max(*result, total); |
| 84 | + return; |
| 85 | + } |
| 86 | + if (total + (i + e) * 120 < *result) { // pruning |
| 87 | + return; |
| 88 | + } |
| 89 | + if (left(n, *curr) || up(n, *curr)) { // leave unoccupied iff left or up is occupied |
| 90 | + curr->emplace_back(0); |
| 91 | + backtracking(m, n, i, e, total, curr, result); |
| 92 | + curr->pop_back(); |
| 93 | + } |
| 94 | + if (i > 0) { |
| 95 | + int new_total = count_total(n, *curr, 1, total); |
| 96 | + curr->emplace_back(1); |
| 97 | + backtracking(m, n, i - 1, e, new_total, curr, result); |
| 98 | + curr->pop_back(); |
| 99 | + } |
| 100 | + if (e > 0) { |
| 101 | + int new_total = count_total(n, *curr, 2, total); |
| 102 | + curr->emplace_back(2); |
| 103 | + backtracking(m, n, i, e - 1, new_total, curr, result); |
| 104 | + curr->pop_back(); |
| 105 | + } |
| 106 | + } |
| 107 | + |
| 108 | + int left(int n, const vector<int>& curr) { |
| 109 | + return (size(curr) % n) ? curr[size(curr) - 1] : 0; |
| 110 | + } |
| 111 | + |
| 112 | + int up(int n, const vector<int>& curr) { |
| 113 | + return (size(curr) >= n) ? curr[size(curr) - n] : 0; |
| 114 | + } |
| 115 | + |
| 116 | + int count_total(int n, const vector<int>& curr, int t, int total) { |
| 117 | + return (total |
| 118 | + - 30 * ((left(n, curr) == 1) + (up(n, curr) == 1)) |
| 119 | + + 20 * ((left(n, curr) == 2) + (up(n, curr) == 2)) |
| 120 | + + (120 - 30 * ((left(n, curr) != 0) + (up(n, curr) != 0))) * (t == 1) |
| 121 | + + ( 40 + 20 * ((left(n, curr) != 0) + (up(n, curr) != 0))) * (t == 2)); |
| 122 | + } |
| 123 | +}; |
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