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.

-2 (K)	-3	3
-5	-10	1
10	30	-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.

  class Solution {
      public int calculateMinimumHP(int[][] dungeon) {
          int m = dungeon.length, n = dungeon[0].length;
          int[][] dp = new int[m][n];
          if(m == 0 || n == 0) return 0;
          dp[m - 1][n - 1] = Math.max(1, 1 - dungeon[m - 1][n - 1]);
          
          for(int i = m - 2; i >= 0; i--){
              dp[i][n - 1]  = Math.max(1, dp[i + 1][n - 1] - dungeon[i][n - 1]); 
          }
          for(int i = n - 2; i >= 0; i--){
              dp[m - 1][i] = Math.max(1, dp[m - 1][i + 1] - dungeon[m - 1][i]);
          }
          for(int i = m - 2; i >= 0; i--){
              for(int j = n - 2; j >= 0; j--){
                  dp[i][j] = Math.max(1, -dungeon[i][j] + Math.min(dp[i+1][j], dp[i][j+1]));
              }
          }
          return dp[0][0];
      }
  }

   

• 

   

   （例子的dp数组）

• 本题重点在于理解到开始点不是0，0而是右下角，是一个往左上逆推的过程。

• 然后就是，dp数组的值最小是1，因为负数不可能完成任务，dp数组是到达当前位置可以剩下的最少血量
• dp数组怎么来的呢？dp[prev_step] + dungeon[prev] = dp[current], 可以逆推dp[prev] == dp[curr] - dungeon[prev].
• 想想，如果dungeon【i】【j】是正的，说明在这个点可以加血，那我dp【i】【j】==1就可以了，秉承最小原则
• 先initialize右下角终点，如果是正数就设置为1，负数就设置为1-（dungeon i，j），再从右往左，下往上initialize最右和最下行列，因为在那里只有一个方向（下/右）
• 最后考虑一般情况，（i，j）可以从右和下得到，选两者小的，因为这样才符合最小值。