【leetcode】Dungeon Game

最新推荐文章于 2019-09-24 14:03:18 发布

Cabinathor

最新推荐文章于 2019-09-24 14:03:18 发布

阅读量260

点赞数

分类专栏： Leetcode

Leetcode 专栏收录该内容

222 篇文章

订阅专栏

本文解析了LeetCode中的一款游戏地图问题，骑士需要通过一个充满威胁与机遇的地图来营救公主。文章详细介绍了如何使用动态规划的方法确定骑士的最小初始健康值，以确保能够成功到达目的地。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

from ： https://leetcode.com/problems/dungeon-game/

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.

从右下角像左上角处理，dp[i][j]表示从坐标(i, j)到右下角所需的血量。初始化时假设最后剩余血里为0，而

状态转移方程：

dungeon[i][j] = max(min(dungeon[i][j+1], dungeon[i+1][j])-dungeon[i][j], 0)

public class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
		int m = dungeon.length;
		int n = dungeon[0].length;
		dungeon[m - 1][n - 1] = max(0 - dungeon[m - 1][n - 1], 0);
		for (int i = m - 2; i >= 0; --i) {
			dungeon[i][n - 1] = max(dungeon[i + 1][n - 1] - dungeon[i][n - 1], 0);
		}
		for (int j = n - 2; j >= 0; --j) {
			dungeon[m - 1][j] = max(dungeon[m - 1][j + 1] - dungeon[m - 1][j], 0);
		}
		for (int i = m - 2; i >= 0; --i) {
			for (int j = n - 2; j >= 0; --j) {
				dungeon[i][j] = max(min(dungeon[i][j + 1], dungeon[i + 1][j]) - dungeon[i][j], 0);
			}
		}
		return dungeon[0][0] + 1;
	}
	
	private int min(int a, int b) {
		if(a < b) {
			return a;
		}
		return b;
	}

	private int max(int a, int b) {
		if(a > b) {
			return a;
		}
		return b;
	}
}

参考：http://www.cnblogs.com/easonliu/p/4237644.html