LeetCode #174. Dungeon Game

迷宫救援：计算骑士最低生命值

最新推荐文章于 2022-11-27 13:55:09 发布

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本文探讨了一个迷宫救援问题，骑士必须从左上角出发，通过右下角救出公主，过程中可能遇到增加或减少生命值的房间。文章提供了一个算法，用于计算骑士救援公主所需的最低初始生命值。

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)

Note:

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) {
        if(dungeon.size()==0||dungeon[0].size()==0) return 1;
        int m=dungeon.size();
        int n=dungeon[0].size();
        // 定义dp[i][j]为骑士到达(i,j)前的生命值，那么最小值为1
        vector<vector<int>> dp(m,vector<int>(n,1));
        // 由于不知道(0,0)处骑士的生命值，可以倒着递推，认为最终骑士的生命值为1
        dp[m-1][n-1]=max(1,1-dungeon[m-1][n-1]);
        for(int i=m-1;i>=0;i--)
        {
            for(int j=n-1;j>=0;j--)
            {
                if(i<m-1&&j<n-1)
                    dp[i][j]=min(dp[i+1][j]-dungeon[i][j],dp[i][j+1]-dungeon[i][j]);
                else if(i<m-1) dp[i][j]=dp[i+1][j]-dungeon[i][j];
                else if(j<n-1) dp[i][j]=dp[i][j+1]-dungeon[i][j];
                dp[i][j]=max(1,dp[i][j]);
            }
        }
        return dp[0][0];
    }
};