Dungeon Game
来源:互联网 发布:xp如何查找网络打印机 编辑:程序博客网 时间:2024/06/06 09:05
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 least7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN
.
拿到题先想到dfs,马上超时。原来dp页可以解决。从尾到头计算到该点需要的最少血量
class Solution {public: int calculateMinimumHP(vector<vector<int> > &dungeon) { int m = dungeon.size(); int n = dungeon[0].size(); vector<vector<int> > dp(m,vector<int>(n)); dp[m-1][n-1] = max(0-dungeon[m-1][n-1],0); for(int i = m-2; i >= 0; i--) dp[i][n-1] = max(0,dp[i+1][n-1]-dungeon[i][n-1]); for(int j = n-2; j >= 0; j--) dp[m-1][j] = max(0,dp[m-1][j+1]-dungeon[m-1][j]); for(int i = m-2; i >= 0; i--)<span class="comment"> //从下向上,从右向左填表</span> for(int j = n-2; j >= 0; j--) dp[i][j] = max(min(dp[i+1][j],dp[i][j+1])-dungeon[i][j], 0); return dp[0][0] +1; }};
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- Dungeon Game
- 获取JSON对象的属性值
- UVA Foreign Exchange(排序)
- gdb如何调试没有符号表(未加-g选项的编译)的程序
- poj 2612 Mine Sweeper
- 从手机丢失看数据安全
- Dungeon Game
- HDU2309 ICPC Score Totalizer Software【水题】
- JSON的parse()和stringfy()方法
- oracle 中 dual 详解
- event.keyCode ,event.which ,event.charCode 键盘事件
- Hive Parquet配置
- android通讯录:拨打电话和发短信
- Android 打造炫目的圆形菜单 秒秒钟高仿建行圆形菜单
- init进程简析