csp-s Analog 92

T1:
  is easy to see that each is independent of the number of
  actually required \ (ax + by = c \ ) solution in \ (| x | + | y | \) minimum
  moguls are calculated directly with different situations minimum
  and konjac I do not count, so I used a strange way:
  briefly modulo find a place close to the minimum value, easy to find \ (| x | + | y | \) is probably a function of a single valley
  so violence can move around to find the bottom xy
 
T2:
  consider a sorting according to the design \ (F_i \) represents the number i of the front and are selected considering the maximum weight of the i-th and the
  first transfer readily occur: \ (F_i = max_ {a_j \ leq b_i} \ {
 
f_j \} + w_i \)   found that the sample can not pass, where considered undercharged
  consideration not be transferred to a section in which i may be added to the sequence
  found when point k satisfies \ (a_k \ geq b_i \ & \ & b_k \
geq a_i \)   it can easily point k i at the back, and without affecting the sequence \ (max \ {a \}
 
\)   then the problem is converted to:
  a sequence, each point has two values \ (Key, Val \) , how to quickly find some interval $ \ sum_ {key_i \ geq C } val_i $
  Chairman of the tree to
 
T3:
  think of the exam binary split multi-source shortest run, but complexity does not seem to
  actually just run it again dijkstra multi-source shortest path, then recorded the shortest path for each point by which particular point expansion of
  the last enumerated answers with information updated each edge of the two end points can be