Finally play all the qvq
T1
30opts
Follow the script is like pressure dp
100opts
We see this as a person and Game
This person is very insidious always give you the worst case
Consider if there are k a k sets assigned to them in just fine
1 may be at least one to fill the k 2
2 may be at least one fill with the k 3
And so on
Also be so long \ (\ sum k ^ {- a_i} \ geq 1 \)
"This question is out to 5e6 but spj have a loj"
T2
7opts
I hung on to write a positive solution
22opts
Vigorously search
42opts
Violence and check collection and
100opts
Since the right interval does not intersect
We can think of it as a cross over the tree (root on the left)
abcabcab
abcabcab
This special case to be sentenced at \ (l2 - l1 <pos - l2 \)
Jump directly to the front so as not to cross repeatedly jump
and so. . We disjoint-set into violence, the complexity of the really right. .
ps if(p && edge[p].l2 <= x && x <= edge[p].r2)here once again sentenced to range
T3
I do not see all sides should be used. . Cancan of
10opts
Repeatedly jump eliminate all lateral sides split into 4:00 Sample Processing
40opts
Direct search
65opts
1. Pruning find the smallest degree point
2. Euler heightening consumer? ? Orz
The orientation of each edge is made into a set of equations or heterologous O (m ^ 3)
(It looks like we have completely get to know)
100opts
For the conventional problem (i.e., without the requirement of even length chain cover)
Can all odd degree twenty-two connected points (virtual edge)
Since the degree of the odd number of points is an even number
So even after the completion of the degree view of all the points are even
On this chart, seeking a Euler then delete all virtual edge
The resulting chain is covered
Now consider the addition of even length limit
Picture necessarily be covered with several chain length of 2 or no solution
Making a chain of length from the rear cover 2
I.e., all the chains of the form (u, p, v) (2 because the length of Well)
Shrunk (u, v)
Because it will shrink the intermediate point (i.e., p) degrees minus 2 degrees do not affect the remaining points
It is still possible to do chain covered by the original method
All chains in the original image obtained at this time is clearly an even length
This question is no solution for the determination of
First, each point of the point is an odd number
Second, the total number of edges must be an even number
The problem has a solution if and only if these two conditions are satisfied
