Leetcode 174. Dungeon Game

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.

思路：典型的二维DP求解，f为状态矩阵，保存当前需要的health。需要注意的是，当f[i][j]为0时，向前推进时为保证knight不死应该为1。比较好的是把f[i][j]扩大一行一列，最后一行一列为int-max。这样就不用初始化计算最后的了。代码会简洁很多。而且，?:运算符与加号连用时需要加括号，不然直接返回赋值了，没经过计算。

class Solution {public:    int calculateMinimumHP(vector<vector<int>>& dungeon) {        int row=dungeon.size();        int col=dungeon[0].size();        vector<vector<int>> f;        f.resize(row);        for(int i=0;i<row;i++) f[i].resize(col,0);                f[row-1][col-1]=dungeon[row-1][col-1]<=0?-dungeon[row-1][col-1]+1:1;                for(int i=row-2;i>=0;i--){            f[i][col-1]=-dungeon[i][col-1]+(f[i+1][col-1]>0?f[i+1][col-1]:1);        }        for(int i=col-2;i>=0;i--){            f[row-1][i]=-dungeon[row-1][i]+(f[row-1][i+1]>0?f[row-1][i+1]:1);        }        for(int i=row-2;i>=0;i--)            for(int j=col-2;j>=0;j--){                f[i][j]=-dungeon[i][j]+(min(f[i+1][j],f[i][j+1])>0?min(f[i+1][j],f[i][j+1]):1);            }        return f[0][0]<=0?1:f[0][0];    }};