LeetCode174—Dungeon Game-蒲公英云

LeetCode174—Dungeon Game

原题

原题链接

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.

分析

题目很有意思，就是在M×N地图上，骑士有N点血，在初始位置(0,0)处，而皇后在(M−1,N−1)的位置，现在地图中每个方格位置(i,j) 都要么能给骑士补血，要么骑士需要扣血，当骑士的血降到小于或等于0，则骑士立刻死亡。问骑士如果要就得皇后，初始状态下至少需要多少血？

骑士向右或者向下走，如果血量小于0就死掉了，这会使得计算变得很复杂。如果我们从后往前看，从最后一个格子逆推回去，就会简单很多。

首先假设dp(i,j)为骑士在(i,j)处的最小血量，那么下个状态，骑士有两个选择：往右，往下。

由于我们是从后往前推，所以现在我们已经知道dp(i+1,j)和dp(i,j+1)。

举个例子：
以dp(i+1,j)为例，假设骑士往下走：
则dungeon(i,j)表示骑士在(i,j)处扣掉的血量或者是补充的血量。

为了完成由(i,j)−>(i+1,j)的状态转移，则必须满足式子：
dp(i,j)+dungeon(i,j)≥dp(i+1,j),dp(i,j)>0

即：
dp(i,j)≥dp(i+1,j)−dungeon(i,j),dp(i,j)>0

要求最小血量，且dp(i,j)必须大于0，因此得出最终的式子：
dp(i,j)=max{ 1,dp(i+1,j)−dungeon(i,j)}

上述式子只考虑一个方向，如果要考虑两个方向，则需考虑取两者较小的值，综合起来得到：
dp(i,j)=min{ max{ 1,dp(i+1,j)−dungeon(i,j)},max{ 1,dp(i,j+1)−dungeon(i,j)}}(1)

根据(1)式写代码就简单了。

代码

#include <iostream>
#include <vector>
#include <string>
#include <iterator>
#include <algorithm>
using namespace std;
class Solution {
    public:
    int calculateMinimumHP(vector<vector<int> >& dungeon) {
        int M = dungeon.size();//row
        int N = dungeon[0].size();//col
        vector<string>path;
        vector< vector<int> >dp(M,vector<int>(N));    
        dp[M-1][N-1]=max(1-dungeon[M-1][N-1],1);
        //last row
        for(int col = N-2;col>=0;--col)
        {
            dp[M-1][col]=max(1,dp[M-1][col+1]-dungeon[M-1][col]);
        }
        //last col
        for(int row = M-2;row>=0;--row)
        {
            dp[row][N-1]=max(1,dp[row+1][N-1]-dungeon[row][N-1]);
        }
        for(int i= M-2;i>=0;--i)
        {
            for(int j=N-2;j>=0;--j)
            {
                int down=max(1,dp[i+1][j]-dungeon[i][j]);
                int right=max(1,dp[i][j+1]-dungeon[i][j]);
                dp[i][j]=min(down,right);
                if(down<right)
                {
                    path.push_back("down");
                }
                else
                {
                    path.push_back("right");
                }
            }
        }
        ostream_iterator<string> output_it(cout);
        adjacent_difference(path.rbegin(),path.rend(),ostream_iterator<string>(cout),
            [](string x,string y)->string{
   cout<<"->";return x;});//打印路径
        cout<<endl;
        return dp[0][0];
    }
};