[DP]174 Dungeon Game

作者: 野生小熊猫 | 来源:发表于2018-10-25 23:26 被阅读0次
    • 分类:DP

    • 考察知识点:DP

    • 最优解时间复杂度:O(n^2)

    • 最优解空间复杂度:O(n^2)

    174. Dungeon Game

    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.

    代码:

    我的方法:

    class Solution:
        def calculateMinimumHP(self, dungeon):
            """
            :type dungeon: List[List[int]]
            :rtype: int
            """
            if len(dungeon)==0 or len(dungeon[0])==0:
                return -1
    
            m=len(dungeon)
            n=len(dungeon[0])
    
            dp=[[float("inf") for i in range(n+1)] for i in range(m+1)]
            dp[m][n-1]=1
            dp[m-1][n]=1
    
            for i in range(m-1,-1,-1):
                for j in range(n-1,-1,-1):
                    dp[i][j]=max(min(dp[i+1][j],dp[i][j+1])-dungeon[i][j],1)
            
    
            return dp[0][0]
    

    讨论:

    1.这道题在Leetcode上被分到了BinarySearch里,我也不知道为啥这么典型的DP题会分到那里
    2.还是推荐花花酱的讲解,很清楚。

    花花大佬的图,思路很清晰 离成为大佬-1s惹

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