LeetCode: Dungeon Game

本文探讨了如何通过动态规划算法计算骑士在穿越迷宫时所需达到的最小初始生命值,以确保能够成功救援被困的公主。通过考虑迷宫中不同房间对骑士生命值的影响,该算法为骑士提供了最优路径选择建议。

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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.
class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon) {
        int x = dungeon.size()-1;
        int y = dungeon[0].size()-1;
        int heal[x+1][y+1];
        heal[x][y] = dungeon[x][y] >= 0 ? 1 : 1-dungeon[x][y];
        for(int i = x-1; i >= 0; i--)
        {
            if(dungeon[i][y] >= 0)
            {
                heal[i][y] = (dungeon[i][y] - heal[i+1][y]) >= 0 ? 1 : heal[i+1][y] - dungeon[i][y];
            }
            else
            {
                heal[i][y] = heal[i+1][y] - dungeon[i][y];
            }
        }
        for(int i = y-1; i >= 0; i--)
        {
            if(dungeon[x][i] >= 0)
            {
                heal[x][i] = (dungeon[x][i] - heal[x][i+1]) >= 0 ? 1 : heal[x][i+1] - dungeon[x][i];
            }
            else
            {
                heal[x][i] = heal[x][i+1] - dungeon[x][i];
            }
        }
        
        for(int i = x-1; i >= 0; i--)
            for(int j = y-1; j >= 0; j--)
            {
                int pre = std::min(heal[i+1][j], heal[i][j+1]);
                if(dungeon[i][j] >= 0)
                {
                    heal[i][j] = (dungeon[i][j] - pre) >= 0 ? 1 : pre - dungeon[i][j];
                }
                else
                {
                    heal[i][j] = pre - dungeon[i][j];
                }                
            }
        
        return heal[0][0];
        
    }
};


资源下载链接为: https://pan.quark.cn/s/22ca96b7bd39 在当今的软件开发领域,自动化构建与发布是提升开发效率和项目质量的关键环节。Jenkins Pipeline作为一种强大的自动化工具,能够有效助力Java项目的快速构建、测试及部署。本文将详细介绍如何利用Jenkins Pipeline实现Java项目的自动化构建与发布。 Jenkins Pipeline简介 Jenkins Pipeline是运行在Jenkins上的一套工作流框架,它将原本分散在单个或多个节点上独立运行的任务串联起来,实现复杂流程的编排与可视化。它是Jenkins 2.X的核心特性之一,推动了Jenkins从持续集成(CI)向持续交付(CD)及DevOps的转变。 创建Pipeline项目 要使用Jenkins Pipeline自动化构建发布Java项目,首先需要创建Pipeline项目。具体步骤如下: 登录Jenkins,点击“新建项”,选择“Pipeline”。 输入项目名称和描述,点击“确定”。 在Pipeline脚本中定义项目字典、发版脚本和预发布脚本。 编写Pipeline脚本 Pipeline脚本是Jenkins Pipeline的核心,用于定义自动化构建和发布的流程。以下是一个简单的Pipeline脚本示例: 在上述脚本中,定义了四个阶段:Checkout、Build、Push package和Deploy/Rollback。每个阶段都可以根据实际需求进行配置和调整。 通过Jenkins Pipeline自动化构建发布Java项目,可以显著提升开发效率和项目质量。借助Pipeline,我们能够轻松实现自动化构建、测试和部署,从而提高项目的整体质量和可靠性。
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