LeetCode 174 Dungeon Game

题目描述

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)

Notes:

  • 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.

一句话题意

一个m×n的矩阵当中,一个骑士从左上角走到右下角,只能向右或向下移动。骑士有初始血量H,矩阵的每一格中有一个数值A[i,j]表示移动到这一格时对血量的增减,计算能保证不在任意时刻血量降至0以下的最小初始血量Hmin

解题思路

棋盘形动归

由题意,在走完最后一个房间的时候血量至少要剩下1,因此最后的状态可以当成是初始状态,由后往前依次决定在每一个位置至少要有多少血量, 这样一个位置的状态是由其下面一个和和左边一个的较小状态决定 .因此一个基本的状态方程是:

dp[i][j] = min(dp[i+1][j], dp[i][j+1]) – dungeon[i][j].

同时要保证每一个状态必须血量都要大于1,因此我们还需要一个方程在保证在每一个状态都大于1,即:

dp[i][j] = max(dp[i][j], 1)

表示虽然当前的血量到最后能够剩下1,但是现在已经低于1了,我们需要为其提升血量维持每一个状态至少都为1

最终的状态转移方程为:

dp[i][j]=max(min(dp[i+1][j],dp[i][j+1])-dungeon[i][j] , 1)

该题可以优化为一位数组,即而其下方的状态可以存储在其当前位置,其状态转移方程为:

dp[i]=max(min(dp[i+1],dp[i])-dungeon[i][j], 1)

 


0 条评论

发表评论