1. This document provides explanations for the AtCoder Regular Contest 038 programming problems in Japanese.
2. It describes the logic and pseudocode for solving a problem about determining if a knight piece on a chessboard will win or lose from its current position. Dynamic programming is used to store and lookup previously computed results.
3. It then provides sample test cases and outputs for the knight problem.
- The document contains code and explanations for solving optimization problems using dynamic programming, including calculating minimum costs using a 2D array to store results.
- It describes applying dynamic programming to problems involving finding minimum costs for tasks that can be split into subtasks, with the overall cost determined by combining subtask costs.
- The code provided shows initializing a 2D array and using nested for loops to iterate through values, calculate minimum costs based on previous results, and store them in the 2D array to build up an optimal solution.
1. This document provides explanations for the AtCoder Regular Contest 038 programming problems in Japanese.
2. It describes the logic and pseudocode for solving a problem about determining if a knight piece on a chessboard will win or lose from its current position. Dynamic programming is used to store and lookup previously computed results.
3. It then provides sample test cases and outputs for the knight problem.
- The document contains code and explanations for solving optimization problems using dynamic programming, including calculating minimum costs using a 2D array to store results.
- It describes applying dynamic programming to problems involving finding minimum costs for tasks that can be split into subtasks, with the overall cost determined by combining subtask costs.
- The code provided shows initializing a 2D array and using nested for loops to iterate through values, calculate minimum costs based on previous results, and store them in the 2D array to build up an optimal solution.
15. 証明
(2)この中でのantichainの最大要素数がmのままの時(m-2個以下になることはない)
P Cの最大のantichainの1つをA’とおく(A’はPでも最大かつA’ ∩ C = ∅であるこ
とに注意)
S = {x∈P | ∃a ∈ A’ : a≦x}, T = {x∈P | ∃a∈A’ : x≦a}とおく
P = S ∪ Tである(S, Tのどちらにも属さないx∈PがあるとするとxはA’に追加するこ
とができA’最大性に反する)
A’⊂SかつA’⊂T
16. 証明
Cの最小元をcとおく
c∈Sと仮定すると, ∃a ∈ A’ : a < cであって, aはCに追加でき
るのでCの最大性に反するのでc∉S
同様にして(Cの最大元)∉Tより|S|, |T| < |P|