Grakn Forces 2020 AF brief solution

$\mathrm{G}\mathrm{r}\mathrm{a}\mathrm{k}\mathrm{n}\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{c}\mathrm{e}\mathrm{s}2020\mathrm{A}-F$ brief solution

Grakn Forces 2020

Note: If the title is multiple sets of data $n$ means$\sum n$

A

Why于题eyes guarantee $a_i\ne b_i\ne c_i$So you can directly determine what to choose bit by bit, the time complexity is $O\left(n\right)$

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B

Greedy addition, until the sequence cannot be added to add a sequence, just simulate it again, the time complexity is $O\left(n\right)$

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C

We consider the bisection time $t$ , and then simulate the left and right end points atttposition after$t$ seconds if$r\le l$ means this$t$ is legal. Time complexity$O(n\log n)$

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D

We consider dp. $f_i$Means to move left $i$ at least move a few spaces up. Finally, take max from back to front. Then$ans=\underset{i=1}{\overset{10^6}{\min }}{f}_i+i$ . Time complexity$O(n\times m)$

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E

We consider the situation that will form a ring, that is, two different numbers $x,y$ appears in two different setsS, TS, Tat the same time$S,T$ species. We consider the elementsxx ineach set$x$ to the current set$S$ connection cost is$a_S+b_x$The side. Then the answer is $(\sum _{i=1}^n\sum _{j=1}^m{a}_i+b_j)-$ The weight of the maximum spanning tree after connecting edges. Because you want to delete the least expensive side. Time complexity$O\left(\right(n+m)\log (n+m))$

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F

Structure questions. First, the two adjacent numbers are equal, and then four are adjacent, and so on.

For example, $n=5:$ 12345 $\sim$ 66345 $\sim$ 66775 $\sim$ 86875 $\sim$ 88885

We just need to divide and conquer to simulate, the time complexity is $O(n\log n)$

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