LeetCode_174: Dungeon Game 动态规划

做了一道很有意思的题目,要求是从左上角开始,在HP不掉到0的情况下,到达右下角。虽然明显是使用动态规划来解,但是其中隐藏了一些技巧 。这里,如果从左上角开始进行DP将使得问题复杂化,技巧就是反向,从终点开始反向DP,这篇博客将详细这种方法的应用条件。


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)



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.






class Solution {
	int calculateMinimumHP(vector<vector<int>>& dungeon) {
		const int m = dungeon.size();
		const int n = dungeon[0].size();
		vector<vector<int>> minHP(m, vector<int>(n, 0));
		minHP[m - 1][n - 1] = max(1, 1-dungeon[m-1][n-1]);
		for (int i = m - 2; i >= 0; --i){
			minHP[i][n-1] = max(1, minHP[i+1][n-1] - dungeon[i][n-1]);

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

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