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| 1 | +Sometimes when we are unable to solve a problem on a tree, we should ask ourselves how to solve it on an array. |
| 2 | + |
| 3 | +There is a DP way to solve this on an array where f(i) is the number of ways |
| 4 | + |
| 5 | +----- |
| 6 | + |
| 7 | +Let us first discuss how to find LIS on an array |
| 8 | + |
| 9 | +We will maintain a DP where f(i) is the minimum element that can end a sequence of length i. |
| 10 | + |
| 11 | +f(i) = oo, if there is no sequence of length i possible. |
| 12 | + |
| 13 | +While processing a new element x, we will look for the first i, such that f(i) >= i and then set f(i) = x |
| 14 | + |
| 15 | +We can do this with binary search to find the LIS in O(N log N) |
| 16 | + |
| 17 | +----- |
| 18 | + |
| 19 | +We will ensure that at each point in time, only those vertices are considered which are on the path from 1 to V. |
| 20 | + |
| 21 | +When we enter a vertex, we will push it into the stack and when we leave it in DFS, we will remove it on the stack. |
| 22 | + |
| 23 | +We have to be able to undo our change to the DP array when we leave a vertex. |
| 24 | +In order to do this, we need to keep track of it's previous states. |
| 25 | + |
| 26 | +We can do this by making dp(i) a stack rather than a single integer. |
| 27 | +When we leave a vertex, we simply pop it from the best dp position for vertex v |
| 28 | + |
| 29 | +We have to modify the binary searches to function over an array of stacks rather than an array of integers |
| 30 | + |
| 31 | +----- |
| 32 | + |
| 33 | +int find_first_greater(int n) |
| 34 | +{ |
| 35 | + int left = 0, right = MAX_N - 1; |
| 36 | + |
| 37 | + //L < x <= R |
| 38 | + while(right - left > 1) |
| 39 | + { |
| 40 | + int mid = (left + right)/2; |
| 41 | + |
| 42 | + if(minimum_ending_of_length[mid].size() == 0 || minimum_ending_of_length[mid].top() >= n) |
| 43 | + { |
| 44 | + right = mid; |
| 45 | + } |
| 46 | + else |
| 47 | + { |
| 48 | + left = mid; |
| 49 | + } |
| 50 | + } |
| 51 | + |
| 52 | + return right; |
| 53 | +} |
| 54 | + |
| 55 | +int find_LIS() |
| 56 | +{ |
| 57 | + int left = 0, right = MAX_N - 1; |
| 58 | + |
| 59 | + //L <= x < R |
| 60 | + while(right - left > 1) |
| 61 | + { |
| 62 | + int mid = (left + right)/2; |
| 63 | + |
| 64 | + if(minimum_ending_of_length[mid].size() > 0) |
| 65 | + { |
| 66 | + left = mid; |
| 67 | + } |
| 68 | + else |
| 69 | + { |
| 70 | + right = mid; |
| 71 | + } |
| 72 | + } |
| 73 | + |
| 74 | + return left; |
| 75 | +} |
| 76 | + |
| 77 | +void dfs(int v, int parent_v) |
| 78 | +{ |
| 79 | + int best_position = find_first_greater(A[v]); |
| 80 | + minimum_ending_of_length[best_position].push(A[v]); |
| 81 | + |
| 82 | + LIS_here[v] = find_LIS(); |
| 83 | + |
| 84 | + for(int child_v : tree[v]) |
| 85 | + { |
| 86 | + if(child_v == parent_v) |
| 87 | + { |
| 88 | + continue; |
| 89 | + } |
| 90 | + |
| 91 | + dfs(child_v, v); |
| 92 | + } |
| 93 | + |
| 94 | + minimum_ending_of_length[best_position].pop(); |
| 95 | +} |
| 96 | + |
| 97 | +int main() |
| 98 | +{ |
| 99 | + int no_of_vertices; |
| 100 | + cin >> no_of_vertices; |
| 101 | + |
| 102 | + for(int i = 1; i <= no_of_vertices; i++) |
| 103 | + { |
| 104 | + cin >> A[i]; |
| 105 | + } |
| 106 | + |
| 107 | + int no_of_edges = no_of_vertices - 1; |
| 108 | + for(int i = 1; i <= no_of_edges; i++) |
| 109 | + { |
| 110 | + int u, v; |
| 111 | + cin >> u >> v; |
| 112 | + |
| 113 | + tree[u].push_back(v); |
| 114 | + tree[v].push_back(u); |
| 115 | + } |
| 116 | + |
| 117 | + dfs(1, 0); |
| 118 | + |
| 119 | + for(int v = 1; v <= no_of_vertices; v++) |
| 120 | + { |
| 121 | + cout << LIS_here[v] << "\n"; |
| 122 | + } |
| 123 | + |
| 124 | + return 0; |
| 125 | +} |
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