天题系列： 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.

要是玩还挺来劲的，想怎做可就费劲了。

ref 思路和code来自http://blog.csdn.net/yangliuy/article/details/42643265

下面是抄人家的思路

“思路分析：这题容易想到的思路是用search DFS或者BFS解，给定起点和终点，我们可以搜索所有从起点到终点的路径，然后贪心保存下来最小的路径权值之和，同时要保证每次扩展分支时当前的生命值状态始终大于0。但是这并不是好的解法，时间复杂度太高。实际上，这题和Unique Path很相似，都是从左上角走到右下角，并且每个格子只能向下或者向右，也就是每个格子只有右边或者下边两个相邻格子。我们可以考虑用动态规划解。关键是如何定义dp数组，这题和Unique Path中dp数组的定义相比有点不同。假设给定的网格是m*n维，我们需要定义dp[][]表示从(i,j)到终点(m-1,n-1)需要的最小生命值，那么递推方程为dp[i][j] = max(min(dp[i][j+1], dp[i+1][j]) - dungeon[i][j], 0)，如果dungeon[i][j]为正，则减去一个正数，初始需要的生命值变小。同理可以理解dungeon[i][j]为负数的情况。和0取最大保证了初始生命值为非负。代码实现很简单，就是初始化后自下而上自右向左填表，典型的二维DP问题的做法。关于这类网格图论问题可以用DFS，BFS，DP以及经典的图算法比如dijkstra 算法，flod算法等等来做，需要多做做这类题目，根据不同情况选择不同算法，增加做这类题目的经验。”

最后返回值+1，比如 input [[0]]

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