LeetCode:174. Dungeon Game

发表于 | 分类于

题目概述

The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.

The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.

Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0’s) or contain magic orbs that increase the knight’s health (positive integers).

In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

Write a function to determine the knight’s minimum initial health so that he is able to rescue the princess.

For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.

-2 (K) -3 3
-5 -10 1
10 30 -5 (P)

Note:

  • The knight’s health has no upper bound.
  • Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

这道题目翻译一下，就是要保证这个骑士在走过每一个格子之后，血量至少为1。这题可以用动态规划来进行求解。如果一个格子上的数字为-5，那么骑士在踏上这个格子的时候，至少需要max(1, 1-(-5)) = 6点血量；如果格子上的数字为5，那么骑士只需要max(1, 1-5) = 1点血量。

接着需要考虑周围格子的情况。如果某个格子的右边的格子至少需要有4点血量，另一个格子至少要3点血量，那么在经过这个格子之后，我们需要至少保留min(4, 3) = 3点血量。

那么，状态转移公式就是dp[i][j] = max(min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j], 1)

代码实现

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
class Solution:
    def calculateMinimumHP(self, dungeon: 'list[list[int]]') -> 'int':
        row = len(dungeon)
        column = len(dungeon[0])
        dp = [[0] * column for _ in range(row)]

        dp[row-1][column-1] = max(1 - dungeon[row-1][column-1], 1)

        for i in range(column-2, -1, -1):
            dp[row-1][i] = max(dp[row-1][i+1] - dungeon[row-1][i], 1)

        for i in range(row - 2, -1, -1):
            dp[i][column - 1] = max(dp[i + 1][column - 1] - dungeon[i][column - 1], 1)

        for i in range(row - 2, -1, -1):
            for j in range(column - 2, -1, -1):
                dp[i][j] = max(min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j], 1)

        return dp[0][0]