[leetcode] 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.
题意：
现在骑士要从左上角去右下角救美丽皇后，救完能发生点什么就靠大家yy了。每一个格子显示的是数字代表骑士到达这一个格子的时候是补充血还是扣血，正的是补充，负的是扣除。我们需要保证骑士能够在前往皇后的路上，到达的每一个格子里血液都要保持是正的，负的或者0就代表跪了，献上了膝盖。
思路：
我们保证骑士到达皇后那里的时候最少得有一滴血，那么他可以靠着这一滴血往下或者往右再走一步，已经不在设危险的格子里了。那么我们倒着考虑路径，骑士可以从dungeon[i+1][j]或者dungeon[i][j+1]到达dungeon[i][j]，那么应该选择一个耗血较少的加上在这一格里面需要耗得血作为总的需求。但是保证血液一定要大于0，如果需求是负的话，那么设为1。
以上。
代码如下：

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