T1:
For the same selection scheme is monotonic in time.
If monotonically decreasing, then the most advantage of this solution at 0, else half position.
Now we need to strike the first m small value at a certain time, you can use nth_element.
Time complexity $ O (nlog \ max (ans)) $.
T2:
And the significance of each path is to the right of the point across.
We can all point to the root fold path, is the weight of all the points with a $ x_1 $.
When asked directly solve the equation, it is determined whether or whether there is an infinite number of solvability solution.
The folding characteristics equation, we need to maintain parity divided by the depth of the right side and the roots can be book-like array.
Increase the number of direct child when you can modify.
Time complexity $ O (nlogn) $.
T3:
To enumerate the right end, the left point to the left from right enumerated, while maintaining a tree array ordinate and the number and prefix.
Ordinate points on both ends divided into many sections, located on the border violence enumerate the interval, then the upper bound of the interval can be obtained lower_bound the lower boundary of the left and right endpoints.
By then the number of points and the prefix in the interval and the calculated contribution can be.
As each point is calculated not more than n times, so the overall complexity of the algorithm is $ O (nmlogm) $.