T1
Subject to the effect
Give you a point right sides are tree 1, find a point such that the distance of the furthest point of the nearest point from the output from the point farthest point to point.
sol
2 except the diameter of the output tree is rounded up.
Proved slightly
T2
Subject to the effect
You \ (n-\) stick and \ (m \) views interrogation, the stick length i \ (a_i \) , each time you ask for two numbers \ (L, R & lt \) , you want to \ (l, \ r) interval is selected to form a triangular stick 3, so that the maximum perimeter. Without a triangle composed of three pieces of wood can be output within the interval -1. \ (1 <= n <= 10 ^ 5,1 <= m <= 5 * 10 ^ 5, a_i <= 10 ^ 9 \)
sol
This correction is the original multi-joint measurement of the title, according to the greedy, before removing the first, second and third long pole, a triangle look can spell, can not continue to remove the second, third and fourth long pole, continues to determine . Worst case stick length Fibonacci number, item more than 45 \ (10 ^ 9 \) , so necessarily determined after 45 makes up a triangle. Now, the question becomes Interval k big problem, since without repair, the Chairman of the tree can be. Time complexity: \ (\ Theta (45mlog_2n) \) .
T3
Subject to the effect
To give you a length \ (n-\) 01 and the string \ (m \) views interrogation, the \ (I \) times will ask you \ (L \) and \ (R & lt \) , you need to output \ (L \) to \ (R & lt \) a maximum value at the end of the longest common prefix of pairwise suffix.
sol
Will not. Positive solutions to use SAM
and LCT
very malignant tumor, may not be able to make up the short term.