Hard-题目43：174. Dungeon Game

题目原文：
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.
题目大意：
给一个二维数组，代表一个网格型的地牢（Dungeon），公主被关在右下角，勇士从左上角开始，一次只能向下或向右走一格，每走一格，他的血量会增加或减少（变化值为数组的对应值），若血量小于等于0，则勇士会死。问要想在最短时间（走最短路径）救出公主，勇士的初始血量至少是多少？
题目分析：
典型的不能再典型的DP，设数组的规模为m*n.
记dp[i][j]为从[I,j]点出发救出公主最少所需血量。首先初始化dp[m][n]=max(1,1-dungeon[m][n]).因为如果右下角血量的变化值为正数，则到此点是加血的，只需血量为1点即可，而如果是负数，则必须保证扣血之后至少剩1点血。
接下来逆向从右下向左上搜索，转移方程dp[i][j] = max(min(dp[i][j+1], dp[i+1][j]) - dungeon[i][j], 0)+1，因为(i,j)点可以从下面来，也可以从右面来，取要求血量最小的一个，再加上扣血值，并保证至少1点血量。如此递推直至dp[1][1]即为所求。
源码：（language：cpp）

class Solution {
public:
    int calculateMinimumHP(vector<vector<int> > &dungeon) {
         if(dungeon.empty())
          return 0;
        vector<vector<int> > dp(dungeon.size());
        for(int i = 0; i < dp.size(); i++) {
            dp[i].reserve(dungeon[i].size());
        }
        int m = dungeon.size();
        int n = dungeon[0].size();
        if(dungeon[m - 1][n - 1] >= 0) {
            dp[m - 1][n - 1] = 1;
        } else {
            dp[m - 1][n - 1] = 1 - dungeon[m - 1][n - 1];
        }

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

        return dp[0][0];
        }

};

成绩：
8ms,100.00%,16ms,65.51%
Cmershen的碎碎念：
这题是Hard后期为数不多的一道还算比较容易想出思路的题，但是比较懒，直接摘抄代码了。