Solutions to some of NOI 2020 questions

foodie

This data range tells us that this problem requires a fast exponentiation of the matrix.

First consider $k=0$ . Divide each point into five points, then the edge weights are all$1$ , and then a Floyd finds the longest way. Transfer is a generalized matrix multiplication, which satisfies the associative law and can be applied to the matrix fast power.

$k\ne 0$ My practice in the examination room is more violent. I directly sort the time of the food festival and ask for violence in sections. The time complexity is$O(125\times n^3\times k\log T)$ , 55 minutes to leave.

Matrix multiplication is an answer matrix multiplied by a transition matrix, because we are from $1$ starts, so the answer matrix only needs the first row. The time complexity is$O(25\times n^2\times k\log T)$。

Making dishes

Did not find $n-2\le$The nature of$m$ , think about the network flow and burst into zero.

Back to the topic, this property tells us that this question is not a network flow. Then the part is divided into $n-1\le m$ , then consider$n-1=$How$m$ is constructed.

Each time the smallest and largest $d$ Take out and make a dish, the smallest$d$ will be used up, and then the largest remaining$d$ Throw it back, which is equivalent to reducing the scale of the problem. You can find that this structure is correct. Use a balanced tree or set to maintain it.

For $n\le m$ , you can choose the largestddeach time$d$ to cook, then throw it back until$n-1\le m$。

For $m=n-2.$ Consider changing$n$ raw materials are divided into two re-sets, if the sum of the two sets is$k$ , then it is equivalent to splitting into two$m=n-1$ situation. This is a typical 01 existence backpack, the complexity is$O\left(n^{2}k\right)$. Because the state is only 01, it can be optimized with bitset, and the time complexity is$O(\frac{n^{2}K}{64})$。