【动态规划】Dungeon Game

标签：c++   程序员   面试   动态规划   

题目：leetcode

Dungeon Game

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.

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.

分析：

设置一个M x N的二维数组f，f[i][j]表示从点[i][j]到右下角，一路上战士血槽的最低点。从右下角一直求到左上角。f的右下角的值是输入数组右下角的值。

之所以要倒着推导（从右下角到左上角），是因为正着推导时（从左上角到右下角），局部最优解不见得是全局最优解。

 int calculateMinimumHP(vector<vector<int> > &dungeon) {
        if(dungeon.empty() || dungeon[0].empty())
        throw exception();
        if(dungeon.size()==1 && dungeon[0].size()==1)
        return dungeon[0][0]>=0?1:dungeon[0][0]*(-1)+1;
        
        int rows = dungeon.size(), cols = dungeon[0].size();
	    vector<vector<int>> f(rows, vector<int>(cols));
    	f.back().back() = dungeon.back().back();
    	
    	for(int i=rows-2;i>=0;i--)
    	{
    	    //f[i][cols-1]=f[i+1][cols-1]+dungeon[i][cols-1];
    	    f[i][cols-1]=min(dungeon[i][cols-1], f[i+1][cols-1]+dungeon[i][cols-1]);
    	}
    		for(int i=cols-2;i>=0;i--)
    	{
    	    //f[rows-1][i]=f[rows-1][i+1]+dungeon[rows-1][i];
    	    f[rows-1][i]=min(dungeon[rows-1][i] ,f[rows-1][i+1]+dungeon[rows-1][i]);
    	}
    	for(int i=rows-2;i>=0;i--)
    	{
    	    for(int j=cols-2;j>=0;j--)
    	    {
    	        //f[i][j]=max(f[i+1][j],f[i][j+1])+dungeon[i][j];
    	        f[i][j]=min(dungeon[i][j] ,max(f[i+1][j],f[i][j+1])+dungeon[i][j]);
    	    }
    	}
    	return f[0][0]>=0?1:1+(-1)*f[0][0];
    }

【动态规划】Dungeon Game

标签：c++   程序员   面试   动态规划   

原文地址：http://blog.csdn.net/bupt8846/article/details/44892699