Dungeon Game -- latched

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.

基本思路：

动态规划

设health[i][j]  为走进dungeon[i][j]的初始血量，且该个血量将能维持到骑士足以走完右下角。

已知条件：骑士走完右下角至少要剩一滴血。即health[m][n-1] = 1。

  m为dungeon行数。

也能够设health[m-1][n] = 1。

此值表示走完右下角的剩余血量。

同一时候也是从该右下角向右，或者向下。走到还有一格时的初始血量。 当然此两格是虚拟的，地牢中不存在。或者形象的说，从右下角向右或者向下走出地牢后，剩余的血量。

从此点。能够倒推出health[0][0]。

骑士仅仅能向右，向下右移动。  要知道当前位置的初始血量，仅仅须要知道其右和其下的初始血量，就能够反推出。

即

health[i][j] = min(health[i+1][j], health[i[j+1])  - dungeon[i][j]

因为骑士要时刻保持血量至少为1. 上面能够改为：

health[i][j] = max(1, min(health[i+1][j], health[i[j+1])  - dungeon[i][j])

因为递推时仅仅须要其右和其下，两个位置， 能够使用滚动数组。用一维替换掉二维数组。

class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon) {
        if (dungeon.empty() || dungeon[0].empty())
            return 0;
        const int m = dungeon.size();
        const int n = dungeon[0].size();
        vector<int> health(n+1, INT_MAX);
        health[n-1] = 1;
        for (int i=m-1; i>=0; i--) {
            for (int j=n-1; j>=0; j--) {
                health[j] = max(1, min(health[j], health[j+1]) - dungeon[i][j]);
            }
        }
        return health[0];
    }
};