zoukankan      html  css  js  c++  java
  • 【leetcode】Dungeon Game

    Dungeon Game

    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.
     
     
    从右下角开始进行动态规划
    设有m行,n列
     
    考虑边界条件:
    最底部的元素:
    若dungeon[m-1][n-1]>=0,则
    dp[m-1][n-1]=0;
    若dungeon[m-1][n-1]<0,则
    dp[m-1][n-1]=-dungeon[m-1][n-1];
    可以简化为dp[m-1][n-1]=max(-dungeon[m-1][n-1],0);
     
    最右边一列
    若dungeon[m-2][n-1]>=0,则
    dp[m-2][n-1]=max(dp[m-1][n-1]-dungeon[m-2][n-1],0);
    若dungeon[m-2][n-1]<0,则
    dp[m-2][n-1]=dp[m-1][n-1]+abs(dungeon[m-2][n-1]);
                          =dp[m-1][n-1]-dungeon[m-2][n-1];
     
     
    化简为dp[m-2][n-1]=max(dp[m-1][n-1]-dungeon[m-2][n-1],0);
     
     
    同理最下边一行
    dp[m-1][n-2]=dp[m-1][n-1]-dungeon[m-1][n-2];
     
     
    对于其他位置的元素
    dp[i][j]=max(min(dp[i+1][j],dp[i][j+1])-dungeon,0)
     
     
     1 class Solution {
     2 public:
     3     int calculateMinimumHP(vector<vector<int> > &dungeon) {
     4        
     5         int m=dungeon.size();
     6         int n=dungeon[0].size();
     7         vector<vector<int>> dp(m,vector<int>(n));
     8        
     9         dp[m-1][n-1]=max(-dungeon[m-1][n-1],0);
    10        
    11         for(int i=m-2;i>=0;i--)
    12         {
    13             dp[i][n-1]=max(dp[i+1][n-1]-dungeon[i][n-1],0);
    14         }
    15        
    16         for(int j=n-2;j>=0;j--)
    17         {
    18             dp[m-1][j]=max(dp[m-1][j+1]-dungeon[m-1][j],0);
    19         }
    20        
    21         for(int i=m-2;i>=0;i--)
    22         {
    23             for(int j=n-2;j>=0;j--)
    24             {
    25                
    26                 dp[i][j]=max(min(dp[i+1][j],dp[i][j+1])-dungeon[i][j],0);
    27             }
    28         }
    29        
    30         return dp[0][0]+1;
    31     }
    32 };
  • 相关阅读:
    2016中国大学生程序设计竞赛
    POJ 2239 化二分图右集合二维为一位的最大匹配
    POJ 1274 二分图最大匹配简单单向
    二分图最大匹配模板
    找割点和割边
    HDU 4432 求因子+进制转换
    HDU 4438 概率 多个情况下的数学期望
    HDU 4424 并查集+贪心思想
    POJ 1611 记录节点数的并查集
    HDU 4430 二分查找
  • 原文地址:https://www.cnblogs.com/reachteam/p/4227382.html
Copyright © 2011-2022 走看看