10.12 Summary

Overview

The total difficulty of completing cf: $2300+2300+2100+2100+2000+2000+900+2100=15800$

Be wyq $\approx 20000$ Hanging

Brief solution

CF1422D

This question 2300 is too much. The portals are sorted directly according to the horizontal and vertical coordinates, and the adjacent sides are violently connected. Then just run the shortest path.

P4552

A good thinking question, consider the interval $[l,r]$ Add and subtract$1$ , which is equivalent to adding and subtracting1 1to the difference array$1$ . So we make the difference array greater than$0$ go to less than$0$ pairs, assuming greater than$0$ is$x$ is$y$ . Then the final difference array must be$(0....,|x-y|)$ So the first answer is$\max (x,y)$ . For the second question, we consider how to set the difference array to$0$ . Then we can change the previous segment$0$ minus$|x-y|$ or subtract by itself, so the number of possible solutions is$|x-y|+1$

CF960E

Consider the tree dp, we set $f_{u,0/1}$It represents $u$ is the number of even/odd numbers.
Then consider the up-and-down discussion in three cases.

1.从uuAn edge is not connected touu in the$u$ subtreeThe points of the$u$ subtree are connected to the edges

2. $In\; the\; u$ subtree, an edge is connected at$u$ subtree is not in$v$ The points of the current subtree are connected to the edges

3. Never at uupoint of$u$ subtree to$u$ subtree inner connecting edge

Dp by situation is fine

CF859E

It is also a classification discussion question. We put a connected block in the front and back position of each person. Consider a connected block. If it is a tree, the answer to the connected block is the size of the block. If the connected block has a ring, it is equivalent to a chain reaction, so there are two cases. If there is a self-loop, then the connected block has one kind. Finally, the multiplication principle can be counted.

CF796D

We consider those edges that can be repeated, which is a point uu of the edge$u$ is covered by two special points so you can delete edges. This bfs can be simulated.

CF771C

It is a tree dp again, we set $f_{x,i}$Represents $x$ distance within subtree$x$ distance$s\%k=i$ the total number of steps,$g_{x,i}$Represents $x$ distance within subtree$x$ distance$s\%k=$The number of points of$i$ .

$f_{x,i}={\sum }_{v\in x}{f}_{v,i-1},g_{x,i}={\sum }_{v\in x}{g}_{v,i-1}$

For $i=1$ us to$f_{x,1}={\sum }_{v\in x}{f}_{v,0}+g_{v,0}$ Because it takes an extra step, these contributions must be added.

Let's first calculate $1$ is the answer to the root, and then change the root dp.

So the answer is ${\sum }_i{\sum }_j{f}_{i,j}$