174. 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.

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.

本题开始的时候做错了，要注意是每个点都要大于0，这道题用dp方法解决，还是自底向上方法解决，当然自顶向下也可以解决。每次每个room还要和1进行比较，其中之前的dp值与当前的-dungeon表示的是残存的血量，而这个残存的血量只保证了到最后的一点时血量剩余1，并不保证每个房间的血量都大于等于1.对于不满足每个房间都大于等于1的dp，要使其值为1.代码如下：

 1 public class Solution {
 2     public int calculateMinimumHP(int[][] dungeon) {
 3         if(dungeon==null||dungeon.length==0||dungeon[0].length==0) return 0;
 4         int m = dungeon.length;
 5         int n = dungeon[0].length;
 6         int[][] dp = new int[m][n];
 7         dp[m-1][n-1] = Math.max(1,1-dungeon[m-1][n-1]);
 8         for(int i=m-2;i>=0;i--){
 9             dp[i][n-1] = Math.max(1,dp[i+1][n-1]-dungeon[i][n-1]);
10         }
11         for(int i=n-2;i>=0;i--){
12             dp[m-1][i] = Math.max(1,dp[m-1][i+1]-dungeon[m-1][i]);
13         }
14         for(int i=m-2;i>=0;i--){
15             for(int j=n-2;j>=0;j--){
16                 int down = Math.max(1,dp[i+1][j]-dungeon[i][j]);
17                 int right = Math.max(1,dp[i][j+1]-dungeon[i][j]);
18                 dp[i][j] = Math.min(down,right);
19             }
20         }
21         return dp[0][0];
22     }
23 }