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Since it need the min value of initial health, this dp tracks from the back to the start.
The trick is either this knight has 1 health or base on next step‘s minimum health minus current dungeon value. Why is minus?
Because when you count the health is health + dungeon[i][j].
1 class Solution { 2 public: 3 int calculateMinimumHP(vector<vector<int> > &dungeon) { 4 if (dungeon.size() == 0) return 1; 5 int n = dungeon.size(), m = dungeon[0].size(); 6 vector<vector<int> > dp(n, vector<int> (m, 0)); 7 dp[n-1][m-1] = max(1, 1 - dungeon[n-1][m-1]); 8 for (int i = n-2; i >= 0; i--) { 9 dp[i][m-1] = max(1, dp[i+1][m-1] - dungeon[i][m-1]); 10 } 11 for (int i = m-2; i >= 0; i--) { 12 dp[n-1][i] = max(1, dp[n-1][i+1] - dungeon[n-1][i]); 13 } 14 for (int i = n-2; i >= 0; i--) { 15 for (int j = m-2; j >= 0; j--) { 16 dp[i][j] = max(1, min(dp[i][j+1], dp[i+1][j]) - dungeon[i][j]); 17 } 18 } 19 return dp[0][0]; 20 } 21 };
LeetCode – Refresh – Dungeon Game
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原文地址：http://www.cnblogs.com/shuashuashua/p/4349403.html
