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:
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3839 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3612 |
7 | Geothermal | 3569 |
8 | ecnerwala | 3494 |
9 | Um_nik | 3396 |
10 | gamegame | 3386 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | Um_nik | 164 |
2 | -is-this-fft- | 162 |
3 | maomao90 | 159 |
3 | atcoder_official | 159 |
5 | cry | 158 |
5 | awoo | 158 |
7 | adamant | 155 |
8 | nor | 154 |
9 | TheScrasse | 153 |
10 | Dominater069 | 152 |
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:
Название |
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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.