LeetCode-Dungeon Game

Dungeon Game

##题目

####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.
    ####Credits:
    Special thanks to @stellari for adding this problem and creating all test cases.

##解题思路
这题容易想到的思路是用search DFS或者BFS解，给定起点和终点，我们可以搜索所有从起点到终点的路径，然后贪心保存下来最小的路径权值之和，同时要保证每次扩展分支时当前的生命值状态始终大于0。但是这并不是好的解法，时间复杂度太高。
该道题由于只能向右和向下移动，因此可以采用动态规划求解，是一个典型的二维动态规划问题，我们可以如下定义：

对于从（0,0）点走到（m-1,n-1)
dp[i][j]:表示从（i,j）走到（m-1.n-1）所需的最小生命值
递推式为：
dp[i][j]=max(min(dp[i+1][j],dp[i][j+1])-dungeon[i][j],0)
如果dungeon[i][j]为正，则减去一个正数，初始需要的生命值变小。同理可以理解dungeon[i][j]为负数的情况。和0取最大保证了初始生命值为非负。

迭代的方法是初始化后，从下往上，从右往左进行填表。

##算法代码
代码采用JAVA实现：

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public class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
        if(dungeon==null)
        	return -1;
        int m=dungeon.length;//行数
        int n=dungeon[0].length;//列数
        int[][] dp=new int[m][n];

        //对m-1行和n-1列进行初始化
        dp[m-1][n-1]=Math.max(0-dungeon[m-1][n-1],0);
        for(int i=n-2;i>=0;i--)
        {
        	dp[m-1][i]=Math.max(dp[m-1][i+1]-dungeon[m-1][i],0);
        }

        for(int i=m-2;i>=0;i--)
        {
        	dp[i][n-1]=Math.max(dp[i+1][n-1]-dungeon[i][n-1],0);
        }

        //进行填表
        for(int i=m-2;i>=0;i--)
        {
        	for(int j=n-2;j>=0;j--)
        	{
        		dp[i][j]=Math.max(Math.min(dp[i+1][j],dp[i][j+1])-dungeon[i][j],0);
        	}

        }

        return dp[0][0]+1;
    }
}

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