zoukankan      html  css  js  c++  java
  • 174. Dungeon Game(动态规划)

    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) -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.
     
     
     
     
     1 class Solution {
     2     public int calculateMinimumHP(int[][] dungeon) {
     3         int n = dungeon.length;
     4         int m = dungeon[0].length;
     5         int[][] dp = new int[n+1][m+1];
     6         for(int i=0;i<=n;i++)
     7             for(int j = 0;j<=m;j++)
     8                 dp[i][j] = Integer.MAX_VALUE;
     9         dp[n][m-1] = 1;
    10         dp[n-1][m] = 1;
    11         
    12         for(int i=n-1;i>=0;i--)
    13             for(int j = m -1;j>=0;j--){
    14                 int need = Math.min(dp[i][j+1],dp[i+1][j]) - dungeon[i][j];
    15                 dp[i][j] = need<=0?1:need;
    16             }
    17         return dp[0][0];
    18     }
    19 }
     
     
     
  • 相关阅读:
    041 Spring Boot中排除功能的处理
    015 图像验证码
    040 Http与RPC
    039 在weblogic下部署jndi的多数据源
    038 lock wait timeout exceeded;try restarting transaction
    037 关于微服务的认识
    036 互联网的框架演变
    035 控制并发 select * from test1 where id =1 for update 就会对这行加锁了?
    034 Maven中的dependencyManagement和dependencies区别
    014 Security的认证流程源码级详解
  • 原文地址:https://www.cnblogs.com/zle1992/p/8996794.html
Copyright © 2011-2022 走看看