LeetCode //C - 174. Dungeon Game

174. Dungeon Game

The demons had captured the princess 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 was initially positioned in the top-left room and must fight his way through 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 (represented by negative integers), so the knight loses health upon entering these rooms; other rooms are either empty (represented as 0) or contain magic orbs that increase the knight's health (represented by positive integers).

To reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

Return the knight's minimum initial health so that he can rescue the princess.

Note that 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.

Example 1:

Input: dungeon = [[-2,-3,3],[-5,-10,1],[10,30,-5]]
Output: 7
Explanation: The initial health of the knight must be at least 7 if he follows the optimal path: RIGHT-> RIGHT -> DOWN -> DOWN.

Example 2:

Input: dungeon = [[0]]
Output: 1

Constraints:
  • m == dungeon.length
  • n == dungeon[i].length
  • 1 <= m, n <= 200
  • -1000 <= dungeon[i][j] <= 1000

From: LeetCode

Link: 174. Dungeon Game


Solution:

Ideas:

1. Dynamic Programming Table (DP Table):

  • Create a 2D array dp where dp[i][j] stores the minimum initial health required to reach the princess starting from cell (i, j).

2. Base Case Initialization:

  • For the princess's cell (m-1, n-1), set dp[m-1][n-1] to the maximum of 1 and 1 - dungeon[m-1][n-1].

3. Filling the DP Table:

  • Last Column: Calculate minimum health for each cell in the last column based on the cell below it.
  • Last Row: Calculate minimum health for each cell in the last row based on the cell to the right.
  • Other Cells: For each cell (i, j), compute the minimum health based on the minimum of the right and below cells, adjusted by the current cell's value. If the resulting health is less than or equal to 0, set it to 1.

4. Result:

  • The value in dp[0][0] represents the minimum initial health required for the knight to rescue the princess starting from the top-left corner.
Code:
c 复制代码
int calculateMinimumHP(int** dungeon, int dungeonSize, int* dungeonColSize) {
    int m = dungeonSize;
    int n = dungeonColSize[0];
    int dp[m][n];
    
    dp[m-1][n-1] = dungeon[m-1][n-1] > 0 ? 1 : 1 - dungeon[m-1][n-1];
    
    for (int i = m - 2; i >= 0; i--) {
        dp[i][n-1] = dp[i+1][n-1] - dungeon[i][n-1];
        if (dp[i][n-1] <= 0) dp[i][n-1] = 1;
    }
    
    for (int j = n - 2; j >= 0; j--) {
        dp[m-1][j] = dp[m-1][j+1] - dungeon[m-1][j];
        if (dp[m-1][j] <= 0) dp[m-1][j] = 1;
    }
    
    for (int i = m - 2; i >= 0; i--) {
        for (int j = n - 2; j >= 0; j--) {
            int min_health_on_exit = dp[i+1][j] < dp[i][j+1] ? dp[i+1][j] : dp[i][j+1];
            dp[i][j] = min_health_on_exit - dungeon[i][j];
            if (dp[i][j] <= 0) dp[i][j] = 1;
        }
    }
    
    return dp[0][0];
}
相关推荐
聚客AI1 天前
🙋‍♀️Transformer训练与推理全流程:从输入处理到输出生成
人工智能·算法·llm
大怪v1 天前
前端:人工智能?我也会啊!来个花活,😎😎😎“自动驾驶”整起!
前端·javascript·算法
惯导马工1 天前
【论文导读】ORB-SLAM3:An Accurate Open-Source Library for Visual, Visual-Inertial and
深度学习·算法
骑自行车的码农1 天前
【React用到的一些算法】游标和栈
算法·react.js
博笙困了1 天前
AcWing学习——双指针算法
c++·算法
moonlifesudo2 天前
322:零钱兑换(三种方法)
算法
NAGNIP2 天前
大模型框架性能优化策略:延迟、吞吐量与成本权衡
算法
美团技术团队2 天前
LongCat-Flash:如何使用 SGLang 部署美团 Agentic 模型
人工智能·算法
Fanxt_Ja2 天前
【LeetCode】算法详解#15 ---环形链表II
数据结构·算法·leetcode·链表
侃侃_天下2 天前
最终的信号类
开发语言·c++·算法