【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.

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 };