I'm interested in whether we can Construct a tree if we know the Starting Time and Ending Time of all the Vertices.
For Example:
- Number of Vertices :4
- Starting time ST[i]:2 4 1 3
- Ending time FT[i]:2 4 4 4
The Tree Looks Like:
# | User | Rating |
---|---|---|
1 | tourist | 3845 |
2 | jiangly | 3707 |
3 | Benq | 3630 |
4 | orzdevinwang | 3573 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | jqdai0815 | 3532 |
8 | ecnerwala | 3501 |
9 | gyh20 | 3447 |
10 | Rebelz | 3409 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 171 |
2 | adamant | 163 |
3 | awoo | 161 |
4 | maroonrk | 152 |
4 | nor | 152 |
6 | -is-this-fft- | 151 |
7 | TheScrasse | 148 |
8 | atcoder_official | 146 |
9 | Petr | 145 |
10 | pajenegod | 144 |
I'm interested in whether we can Construct a tree if we know the Starting Time and Ending Time of all the Vertices.
For Example:
The Tree Looks Like:
Name |
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There exists a unique tree for each VALID start and end time arrays.
A small hint for the tree construction : The direct parent of a node i shall be a node j with maximum start time, following the constraint that start[j] < start[i] and end[j] ≥ end[i].
There exists such a node j for each node i, except the root.
A node having s_t == f_t must be leaf node. First insert the node having s_t == 1 into the tree and set pos(a variable) equal to this node. This "pos" will tell u the the position where to append the next_node into the tree. Along with it maintain a parent array. After appending a node(say i) at pos, check if s_t[i] == f_t[i], If yes then backtrack to the parent until u find a parent having f_t greater than the s_t of node i. This will become the new pos pointer. And if "no" then pos will be set to i for next node.In this way u can insert all nodes.