[leetcode]174. 地下城游戏

写了半天从[0][0]开始的，代码有90多行，就过了一个样例。。看了题解，原来只能倒着来。。额额

1、题解里写得最简洁的：

https://leetcode-cn.com/problems/dungeon-game/solution/c-dp-10xing-by-pjpj/

dp[i][j]表示从[i][j]位置到[m-1][n-1]需要的最小血量，即到[i,j]时勇士的血量最少得这么多

dp[i][j] = max(1, (min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j]));

class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon) {
        int m = dungeon.size(), n = dungeon[0].size();
        vector<vector<int>>dp(m + 1, vector<int>(n + 1, INT_MAX/10)); //初始化为无穷大
        dp[m][n-1] = 1;
        for(int i = m - 1; i >= 0; i--)
        {
            for(int j = n - 1; j >= 0; j--)
            {
                dp[i][j] = max(1, (min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j]));
            }
        }
        return dp[0][0];
    }
};

2、这个可能好理解点

class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon) {
        int m = dungeon.size(), n = dungeon[0].size();
        vector<vector<int>>dp(m, vector<int>(n, 0));

        //当dungeon[m-1][n-1] >= 0,这 1 - dugeon[m-1][n-1] <= 1
        //反之,1 - dugeon[m-1][n-1] > 1
        dp[m-1][n-1] = max(1, 1 - dungeon[m-1][n-1]);

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

3、空间优化：

class Solution {
public:
    int calculateMinimumHP(vector<vector<int>>& dungeon) {
        int m = dungeon.size(), n = dungeon[0].size();
        vector<int>dp(n + 1, INT_MAX/10);
        dp[n-1] = 1;
        for(int i = m - 1; i >= 0; i--)
        {
            dp[n-1] = max(1, dp[n-1] - dungeon[i][n-1]);
            for(int j = n - 2; j >= 0; j--)
            {
                dp[j] = max(1, (min(dp[j], dp[j+1]) - dungeon[i][j]));
            }
        }
        return dp[0];
    }
};