174. Dungeon-Game

difficulty: Hard

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

table.dungeon, .dungeon th, .dungeon td { border:3px solid black; } .dungeon th, .dungeon td { text-align: center; height: 70px; width: 70px; }

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.

Method One

class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
        int m = dungeon.length;
        int n = dungeon[0].length;
        int[][] dp = new int[m][n];
        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 j = n - 2; j >= 0; j--){
            dp[m - 1][j] = Math.max(1, dp[m - 1][j + 1] - dungeon[m - 1][j]);
        }
        for(int i = m - 2; i >= 0; i-- ){
            for(int j = n - 2; j >= 0; j-- ){
                int hp = Math.min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j];
                dp[i][j] = Math.max(1, hp);
            }
        }
        return dp[0][0];
    }
}
​
/**
    这个问题比较反直觉的是，它只能用终点反推起点。
    为什么？因为要考虑什么才是这个问题的子问题。
    对于子问题 我们不能够改变的就是这个终点 终点变了问题就变了
    因此我们只能够从终点倒推到某个起点
    二维数组 dp[i][j] 储存的就是我们的所有子问题的答案，即
    从 i,j 出发到终点所需要的最小体力
    状态转移 dp[i][j] = Math.min(dp[i + 1][j], dp[i][ j + 1]) - dungeon[i][j];
    如果数字为负，说明我们这里是加体力的，我们只需要一个 +1 的体力走到这里就行了。
*/
​