174. Dungeon Game

题目：

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.

解答：
这一题最重要的是把 ： 所剩血量 ＝ 初始化血量＋走这一步消耗的血量 >= 1 这句话读懂。那么我们假设f(i, j)是走完(i, j)后所剩余的血量（f(i, j)肯定是大于等于1的）。如果想存活下来，最少需要f(i, j) ＝ f(上一步血量）＋ dungeon(i, j) >= 1，即：f(i, j) = Math.max(1, f(上一步血量）- dungeon(i, j)). 然后分类讨论上一步血量的可能性，注意边界情况的初始化即可。

程序如下：

public class Solution {
    //State: f[i][j] is from (0, 0) to (i, j) the least health the knight should have
    //Function: f[i][j] + dungeon[i][j] >= 1 --> f[i][j] = Math.max(1 - dungeon[i][j], 1);
    //Intialize: f[i][n - 1] = Math.max(1 - f[i + 1][n - 1], 1), f[m - 1][i] = math.max(i - f[m - 1][i + 1], 1);
    //Result: f[0][0];
    public int calculateMinimumHP(int[][] dungeon) {
        if (dungeon == null || dungeon.length == 0 || dungeon[0].length == 0) {
            return 0;
        }
        
        int m = dungeon.length, n = dungeon[0].length;
        int[][] f = new int[m][n];
        //初始化最后一步的血量要求
        f[m - 1][n - 1] = Math.max(1, 1 - dungeon[m - 1][n - 1]);
        //最后一列格子的初始血量
        for (int i = m - 2; i >= 0; i--) {
            f[i][n - 1] = Math.max(1, f[i + 1][n - 1] - dungeon[i][n - 1]);
        }
        //最后一排格子的初始血量
        for (int j = n - 2; j >= 0; j--) {
            f[m - 1][j] = Math.max(1, f[m - 1][j + 1] - dungeon[m - 1][j]);
        }
        //可以从右边或者下边得到当前格子的最小初始血量
        for (int i = m - 2; i >= 0; i--) {
            for (int j = n - 2; j >= 0; j--) {
                int right = Math.max(1, f[i][j + 1] - dungeon[i][j]);
                int down = Math.max(1, f[i + 1][j] - dungeon[i][j]);
                f[i][j] = Math.min(right, down);
            }
        }
        
        return f[0][0];
    }
}