一. 題目描述

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.

二. 題目分析

題目很長，大致介紹了一款遊戲的規則：

惡魔抓走了公主(P)並把她囚禁在地牢的右下角（矩陣圖中的(P)所在的格子）。地牢包含M * N個房間。騎士(K)一開始位於左上角的格子裡，目標是拯救公主。

騎士擁有初始生命值，為一正整數。如果在任何一個格子中，生命值變為0或小於0，他就會掛掉。

一些房間由惡魔守衛著，因此當騎士進入這些房間時就會損失生命值（負整數）；其他房間有兩種，一種是空房間（數值為0），另外一種房間是存在一些魔力寶石，這些寶石可以增加騎士的生命值（正整數）。

為了儘快的解救到公主，騎士決定每一步只向右或者向下移動。

編寫一個函式決定騎士的最小初始生命值，確保他可以成功營救公主。

題目給出一個例子，該例子中騎士的初始生命值至少應當為7，如果他按照下面的最優路線行進：右 -> 右 -> 下 -> 下.

該題目描述非常複雜，其實看懂規則後直接看例子比較容易理解，可使用動態規劃來解決。令dungeon[i][j]表示從座標(i, j)所在的格子出發，到最右下角公主所在處所需的最小血量。寫出狀態轉換方程：

dungeon[i][j] = max(min(dungeon[i][j + 1], dungeon[i + 1][j]) - dungeon[i][j], 0)

三. 示例程式碼

class Solution {
public:
    int calculateMinimumHP(vector<vector<int> > &dungeon) {
        int m = dungeon.size();
        int n = dungeon[0].size();
        dungeon[m - 1][n - 1] = max(1 - dungeon[m - 1][n - 1], 1);
        // 邊界值初始化
        for (int i = m - 2; i >= 0; --i) 
            dungeon[i][n - 1] = max(dungeon[i + 1][n - 1] - dungeon[i][n - 1], 1);

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

        return dungeon[0][0];
    }
};

四. 小結

該題有一定的難度，但如果能推出狀態轉換公式，使用動態規劃可快速解決。