Dp am talking about digital and knapsack problem.
Talk about backpack:
Full backpack: I changed the order:
Multiple backpack:
Multiple backpack Optimization:
So put each item into these groups, then put them into different items, it becomes a 01 knapsack problem;
Sliding window value takes the most problems. Monotonous queue optimization.
The method is simple, wherein in each group can be enumerated a computing article.
Tips:
Some ignorant. . .
:
Finally, to the digital link :( dp nausea the morning.)
dp method:
Determine the upper bound.
If we want to enumerate to 2147, currently it has to enumerate the second place, if the enumeration to 1, then we say that he has reached the upper bound, the next one can enumerate from 0 to 4. If this bit is 0, the next one is due no matter how much this number will never be greater than the original number, it can be any enumerated from 0-9. After a few as well.
in the afternoon:
Then say the tree dp:
end