Codeforces Round #677 (Div. 3, CF1433) solution

Solution

T1

Direct violence is enough. Mathematical techniques can also be used.

T2

For every two consecutive segments $An$ interval of00The sum of$0$ is the answer.

T3

First, if all numbers are the same, obviously output $-1$。

Otherwise, there must be a maximum that can eat all other numbers. We only need to find a maximum value so that its left side or its right side are not all maximum values.

Time complexity $O\left(n\right)$。

T4

An ingenious and somewhat watery structure problem.

First, we sort the entire sequence. If all numbers in this sequence are the same, obviously it won't work; otherwise, we find the longest prefix with equal numbers . Then, the first number is connected to all numbers not in this prefix , and the last number is connected to all numbers in this prefix except the first number .

Time complexity $O\left(n\right)$ . That$n\le 5000$ There is a bit of water.

T5

A combinatorial math problem that compares water. Let $m=\frac{n}{2}$。

First, there is a total of $A_n^m\times m!$ Kinds of plans (if the same plan is not deleted).

It can be found that { 1, 3 1,3 $1,3$}{ $2,4,5$ } and { 2, 4, 5 2,4,5 $2,4,5$}{ $1,3$ } Essentially the same, so divide by$2$。

Adhesion, { 1, 3, 4 1,3,4 $1,3,4$ } and { 3, 4, 1 3,4,1 $3,4,1$}, { $4,1,3$ } are essentially the same. For a set there are$\frac{m}{2}$Kind, but there are two sets, so divide by $(\frac{m}{2}{)}^2$。

So the answer is $$\frac{A_n^m\times m!}{2\times (\frac{m}{2}{)}^2}$$。

Why can't even Ni Yuan take the exam? Just make a $n\le 20$ ? Is there any distinction...

Trivia: This problem can also be solved with hash!

T6

A dp dp with a little bit of thinking content$dp$ title.

First, we run dp dp for each row$dp$ . The state is designed as$d{p}_{i,j}$, Which means that in this row, a total of ii is selectedThe number of$i$ , the sum of these numberskkThe value of$k$ is$When\; j$ , the maximum value of the sum of these numbers.

Then, for each row, for all $i\le \frac{m}{2}$, We all save all $d{p}_{i,j}$. Then, for multiple rows, we run the grouping backpack again and it's done.

Time complexity $A\left(nmk\right)$ .

T7

A good graph topic.

First, we first run out all $x,$The shortest path between$y$ (using$Dijkstra$ ), and then enumerate the edges (that is, change the weight of this edge to$0$ ). Suppose this edge is$u->v$ , then if$x$到$If$ the shortest path of$y$ changes, it must become:

①From $x$ arrival$u$， 来$u$ to$v$ from$v$ to$y$ ;
②From$x$ to$v$ from$v$ arrival$u$， 来$u$到$and$。

It can be found that we only need to run out with $u,v$ is the shortest path of the source, and then try to decrease$x$到yyThe shortest path of$y$ is fine. Since this is an undirected graph, the correctness of this solution is guaranteed (in an undirected graph,$a$ tobbThe shortest path of$b$ is$b$ to$a$ shortest path)

Time complexity $O\left(\right(n+m)(k+m\left)logm\right)$ , you can pass this question.

Code

Cuckoo