def quick_sort(arr, low, high, k):
if low < high:
pivot_index = partition(arr, low, high, k)
quick_sort(arr, low, pivot_index - 1, k)
quick_sort(arr, pivot_index + 1, high, k)


def partition(arr, low, high, k):
pivot = arr[high][k]
i = low - 1
for j in range(low, high):
if arr[j][k] <= pivot:
i += 1
arr[i], arr[j] = arr[j], arr[i]
arr[i + 1], arr[high] = arr[high], arr[i + 1]
return i + 1


def determine_final_hp(playerU, playerD):
for i in range(len(playerD)):
for j in range(len(playerU)):
k = len(playerU) - j - 1
playeri = playerD[i]
playerj = playerU[k]
if playeri[0] >= playerj[0] and playeri[1] > 0 and playerj[1] > 0:
if playeri[1] < playerj[1]:
playeri[1] = 0
playerj[1] -= 1
elif playeri[1] > playerj[1]:
playerj[1] = 0
playeri[1] -= 1
else:
playeri[1] = 0
playerj[1] = 0
return playerU, playerD


n = int(input())
playersU = []
playersD = []
players = []
for i in range(n):
floor, hp, direction = input().split()
if direction == ‘U’:
playersU.append([int(floor), int(hp), i])
else:
playersD.append([int(floor), int(hp), i])
quick_sort(playersU, 0, len(playersU) - 1, 0)
quick_sort(playersD, 0, len(playersD) - 1, 0)
playersU, playersD = determine_final_hp(playersU, playersD)
for u in range(len(playersU)):
players.append(playersU[u])
for d in range(len(playersD)):
players.append(playersD[d])
quick_sort(players, 0, len(players) - 1, 2)
for i in range(len(players)):
player = players[i]
if player[1] != 0:
print(player[1])按下述内容，改进它使它在面对百万级的n时也能有较好的表现
The current implementation of the code is not efficient when dealing with large input sizes. Here are a few improvements that can be made to optimize the code:

1. Use a more efficient sorting algorithm: The current implementation uses the Quick Sort algorithm. While Quick Sort is generally fast, it has a worst-case time complexity of O(n^2) which will be inefficient for large input sizes. Consider using a more efficient sorting algorithm like Merge Sort or Heap Sort which have a worst-case time complexity of O(nlogn).

2. Use a better pivot selection strategy: The current implementation always selects the last element as the pivot. This can lead to inefficient partitioning, especially when the input is already partially sorted or reverse sorted. Consider using a better pivot selection strategy like choosing the median of three random elements to minimize the chance of inefficient partitioning.

3. Avoid unnecessary comparisons: In the determine_final_hp function, nested loops are used to compare all player combinations. It is possible to optimize this by avoiding unnecessary comparisons. For example, once a player from playersD defeats a player from playersU, there is no need to continue checking that defeated player against the remaining players from playersU. This can be achieved by breaking out of the inner loop once a defeat occurs.

4. Use a data structure to store player indices instead of sorting an entire array: Instead of sorting the entire players array using the third index, you can use a data structure like a priority queue or a balanced binary search tree to efficiently keep track of the indices while sorting.

By implementing these improvements, the code will have better performance and should be able to handle larger input sizes more efficiently.