[Leetcode] 174. Dungeon Game 解题报告

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)-33-5-1011030-5 (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.

思路：

一道典型的动态规划题目。不过和其它动规题目不同的是，我们从右下角往左上角倒推更加容易。一个位置的状态是由它下面的一个和右边的状态决定。因此一个基本的状态方程是：dp[i][j] = min(max(1, dp[i + 1][j] - dungeon[i][j]), max(1, dp[i][j + 1] - dungeon[i][j]))。

代码：

1、O(n^2)空间复杂度的代码：

class Solution {public:    int calculateMinimumHP(vector<vector<int>>& dungeon) {        if(dungeon.size() == 0 || dungeon[0].size() == 0) {            return 0;        }        int row = dungeon.size(), col = dungeon[0].size();        vector<vector<int>> dp(row, vector<int>(col, 1));        dp[row - 1][col - 1] = max(1, 1 - dungeon[row - 1][col - 1]);        for(int i = row - 1; i >= 0; --i) {            for(int j = col - 1; j >= 0; --j) {                if(i == row - 1 && j == col - 1) {    // The initical case                    continue;                }                int min_from_row = INT_MAX;                int min_from_col = INT_MAX;                if(i < row - 1) {                    min_from_col = max(1, dp[i + 1][j] - dungeon[i][j]);                }                if(j < col - 1) {                    min_from_row = max(1, dp[i][j + 1] - dungeon[i][j]);                }                dp[i][j] = min(min_from_row, min_from_col);            }        }        return dp[0][0];    }};

2、O(n)空间复杂度的代码：

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