leetcode-174. Dungeon Game

本文介绍了一个利用动态规划算法解决的游戏问题——骑士以最少初始血量成功救出公主。通过从终点反向推导至起点的方式,确保了每一步都是向着最优解迈进。
class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon){
        //正着看的话,只有局部最优,没有全局最优,类似于贪心
        //应该倒着动态规划
        int m = dungeon.size();
        int n = dungeon[0].size();
        vector<vector<int>> dp(m+1, vector<int>(n+1, INT_MAX));//初始化要为最大值!!!
        //dp中存储在此位置所需要的最小的体力值
        dp[m][n-1] = 1;
        dp[m-1][n] = 1;
        for(int i=m-1;i>=0;i--) //i往上走
            for(int j=n-1;j>=0;j--){ //j往左走
                int need = min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j];
                dp[i][j] = need <= 0 ? 1 : need;
            }
        //体力不能等于0,至少要为1,所以need最小是1
        return dp[0][0];
    }
};

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 (positiveintegers).

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)-33
-5-101
1030-5 (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.

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值