結果
| 問題 |
No.126 2基のエレベータ
|
| ユーザー |
 qwewe
|
| 提出日時 | 2025-05-14 12:57:05 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 5,892 bytes |
| コンパイル時間 | 144 ms |
| コンパイル使用メモリ | 82,632 KB |
| 実行使用メモリ | 54,356 KB |
| 最終ジャッジ日時 | 2025-05-14 12:58:57 |
| 合計ジャッジ時間 | 2,343 ms |
|
ジャッジサーバーID （参考情報） |
judge4 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 23 WA * 4 |
ソースコード
import sys
# Function to read input and calculate the minimum total elevator travel distance
def solve():
# Read the initial positions of elevator A, elevator B, and Wilson (S)
# A: position of elevator A (can be 0 to 100, where 0 is B1)
# B: position of elevator B (can be 1 to 100)
# S: Wilson's starting position (must be 1 to 100)
# Floor 0 represents B1 (Basement 1)
a, b, s = map(int, sys.stdin.readline().split())
# --- Case 1: Wilson is at floor 1 (S=1) ---
# Wilson needs to go down to floor 0 (B1).
# The problem states: "if the down button is pressed on the 1st floor (S=1), ... only elevator A will come".
# This rule applies regardless of elevator B's position or distance.
if s == 1:
# Calculate the total cost for elevator A to serve Wilson:
# 1. Elevator A moves from its current position 'a' to floor 1.
# If A is already at floor 1 (a=1), this movement cost is 0.
# Otherwise, the cost is the distance |a - 1|.
cost_a_to_1 = abs(a - 1)
# 2. Wilson gets into elevator A at floor 1.
# 3. Elevator A moves from floor 1 to floor 0 (B1).
# The cost is the distance |1 - 0| = 1.
cost_1_to_0 = 1
# The total cost is the sum of these two movements by elevator A.
total_cost = cost_a_to_1 + cost_1_to_0
print(total_cost)
# We have found the answer for S=1, so we can exit.
return
# --- Case 2: Wilson is at floor S > 1 ---
# Wilson needs to call an elevator to floor S first.
# Subcase 2.1: Elevator A is already at Wilson's floor S (a == s)
# The rule "If an elevator is already at the floor, ... the door just opens" applies.
if a == s:
# Elevator A opens its doors (0 movement cost for pickup).
# Wilson gets into elevator A.
# Elevator A moves directly from floor S to floor 0 (B1).
# The cost is the distance |s - 0| = s (since s > 1 >= 0).
total_cost = s
print(total_cost)
# We have found the answer, so we can exit.
return
# Subcase 2.2: Elevator B is already at Wilson's floor S (b == s), and Elevator A is not (a != s)
# The "already there" rule applies for elevator B.
if b == s:
# Elevator B opens its doors (0 movement cost for pickup).
# Wilson gets into elevator B.
# Since elevator B cannot go to floor 0, Wilson must take B to a floor accessible by A.
# The most logical floor is floor 1, as it's the closest floor to 0 that B can reach.
# 1. Elevator B moves from floor S to floor 1.
# The cost is the distance |s - 1| = s - 1 (since s > 1).
cost_b_s_to_1 = s - 1
# Wilson gets off elevator B at floor 1.
# Now, Wilson needs elevator A to go from floor 1 to floor 0.
# Wilson calls elevator A (using the Down button at floor 1, which guarantees A comes).
# Elevator A is currently at its initial position 'a'.
# 2. Elevator A moves from position 'a' to floor 1.
# The cost is the distance |a - 1|.
cost_a_to_1 = abs(a - 1)
# Wilson gets into elevator A at floor 1.
# 3. Elevator A moves from floor 1 to floor 0.
# The cost is the distance |1 - 0| = 1.
cost_a_1_to_0 = 1
# The total cost is the sum of movements by both elevator B and elevator A.
# total_cost = cost_b_s_to_1 + cost_a_to_1 + cost_a_1_to_0
# Simplified: total_cost = (s - 1) + abs(a - 1) + 1 = s + abs(a - 1)
total_cost = s + abs(a - 1)
print(total_cost)
# We have found the answer, so we can exit.
return
# Subcase 2.3: Neither elevator A nor B is at Wilson's floor S (a != s and b != s)
# Wilson presses a button at floor S. The elevator rules determine which one comes.
# Rule: The closer elevator comes. If distances are equal, A is preferred.
# Calculate the distance from each elevator to floor S.
dist_a = abs(a - s)
dist_b = abs(b - s)
# Compare distances to determine which elevator responds.
if dist_a <= dist_b:
# Elevator A comes because it's closer or the distances are equal.
# 1. Elevator A moves from position 'a' to floor S. Cost = dist_a.
cost_a_to_s = dist_a
# Wilson gets into elevator A at floor S.
# 2. Elevator A moves from floor S to floor 0. Cost = |s - 0| = s.
cost_s_to_0 = s
# Total cost is the sum of movements by elevator A.
total_cost = cost_a_to_s + cost_s_to_0
print(total_cost)
# End of this path
else: # dist_a > dist_b
# Elevator B comes because it is strictly closer.
# 1. Elevator B moves from position 'b' to floor S. Cost = dist_b.
cost_b_to_s = dist_b
# Wilson gets into elevator B at floor S. Takes B to floor 1.
# 2. Elevator B moves from floor S to floor 1. Cost = |s - 1| = s - 1.
cost_b_s_to_1 = s - 1
# Wilson gets off at floor 1. Calls elevator A (still at initial position 'a').
# 3. Elevator A moves from position 'a' to floor 1. Cost = |a - 1|.
cost_a_to_1 = abs(a - 1)
# Wilson gets into elevator A at floor 1.
# 4. Elevator A moves from floor 1 to floor 0. Cost = |1 - 0| = 1.
cost_a_1_to_0 = 1
# Total cost is the sum of movements by both B and A.
# total_cost = cost_b_to_s + cost_b_s_to_1 + cost_a_to_1 + cost_a_1_to_0
# Simplified: total_cost = dist_b + (s - 1) + abs(a - 1) + 1 = dist_b + s + abs(a - 1)
total_cost = dist_b + s + abs(a - 1)
print(total_cost)
# End of this path
# Execute the solve function to read input and print the result
solve()