LeetCode刷题笔录dungeon game_leetcode optimal path-优快云博客

探讨如何确定骑士从迷宫左上角出发，仅向右或向下移动到达右下角并救援公主所需的最少初始生命值。面对含有威胁与补给的房间，通过动态规划算法找出最优路径。

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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.

挺有意思的一道题，基本上可以一眼看出要用DP。问题就是DP的递归方程怎么写。

dp[i][j]表示进入这个格子后保证knight不会死所需要的最小HP。如果一个格子的值为负，那么进入这个格子之前knight需要有的最小HP是-dungeon[i][j] + 1.如果格子的值非负，那么最小HP需求就是1.

这里可以看出DP的方向是从最右下角开始一直到左上角。首先dp[m-1][n-1] = Math.max(1, -dungeon[m-1][n-1] + 1).

递归方程是dp[i][j] = Math.max(1, Math.min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j]).

public class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
        if(dungeon.length == 0)
            return 0;
        int m = dungeon.length, n = dungeon[0].length;
        int[][] dp = new int[m][n];
        dp[m - 1][n - 1] = dungeon[m - 1][n - 1] >= 0 ? 1 : -dungeon[m - 1][n - 1] + 1;
        for(int i = m - 2; i >= 0; i--){
            dp[i][n - 1] = Math.max(dp[i + 1][n - 1] - dungeon[i][n - 1], 1);
        }
        for(int j = n - 2; j >= 0; j--){
            dp[m - 1][j] = Math.max(dp[m - 1][j + 1] - dungeon[m - 1][j], 1);
        }
        
        for(int i = m - 2; i >= 0; i--){
            for(int j = n - 2; j >= 0; j--){
                dp[i][j] = Math.max(1, Math.min(dp[i + 1][j], dp[i][j + 1]) - dungeon[i][j]);
            }
        }
        return dp[0][0];
    }
}