QUESTION

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)

1st TRY

class Solution {
public:
int calculateMinimumHP(vector<vector<int> > &dungeon) {
minInitHP = INT_MAX;
dfs(dungeon,,,,);
return minInitHP+1;
}
void dfs(vector<vector<int> > &dungeon, int m, int n, int currentHP, int currentMinInitHP)
{
currentHP -= dungeon[m][n];
currentMinInitHP= max(currentMinInitHP,currentHP);
if(currentMinInitHP>minInitHP) return;
if(m==dungeon.size()- && n==dungeon[].size()-)
{
currentHP = min(currentHP,currentMinInitHP);
return;
}
if(m!=dungeon.size()-) dfs(dungeon,m+,n,currentHP,currentMinInitHP);
if(n!=dungeon[].size()-) dfs(dungeon,m,n+,currentHP,currentMinInitHP); }
private:
int minInitHP;
};

Result: Time Limit Exceeded

2nd TRY

用空间换时间,动态规划

class Solution {
public:
int calculateMinimumHP(vector<vector<int> > &dungeon) {
int m = dungeon.size();
int n = dungeon[].size(); int **dp = new int*[m]; // min HP needed when knight come here
for(int i=; i < m; i++)
dp[i] = new int[n]; dp[][] = - dungeon[][];for(int i = ; i < m; i++)
{
dp[i][] = max(dp[i-][]-dungeon[i][], dp[i-][]);
}
for(int i = ; i < n; i++)
{
dp[][i] = max(dp[][i-]-dungeon[][i], dp[][i-]);
}
for(int i = ; i < m; i++)
{
for(int j = ; j < n; j++)
{
dp[i][j]=min(dp[i-][j],dp[i][j-])-min(,dungeon[i][j]);
}
}
return max(,dp[m-][n-]+);
}
};

Result: Wrong Answer

Input: [[0,5],[-2,-3]]
Output: 4
Expected: 1

3rd TRY

状态的保存有问题,需要保存两个状态:到当前格在初始时需要的HP,以及到了当前格的HP

所以要从下往上填状态,那么只要保存一个状态,当前格的HP

class Solution {
public:
int calculateMinimumHP(vector<vector<int> > &dungeon) {
int m = dungeon.size();
int n = dungeon[].size(); int **dp = new int*[m]; // min HP needed when knight from here to end
for(int i=; i < m; i++)
dp[i] = new int[n]; dp[m-][n-] = max( - dungeon[m-][n-],);
for(int i = m-; i >= ; i--)
{
dp[i][n-] = max(dp[i+][n-]-dungeon[i][n-],);
}
for(int i = n-; i >= ; i--)
{
dp[m-][i] = max(dp[m-][i+]-dungeon[m-][i],);
}
for(int i = m-; i >= ; i--)
{
for(int j = n-; j >= ; j--)
{
dp[i][j]=max(min(dp[i+][j],dp[i][j+])-dungeon[i][j],);
}
}
return dp[][]+;
}
};

Result: Accepted

Dungeon Game (GRAPH - DP)的更多相关文章

  1. SCOJ 4427: Miss Zhao's Graph dp

    4427: Miss Zhao's Graph 题目连接: http://acm.scu.edu.cn/soj/problem.action?id=4427 Description Mr Jiang ...

  2. F. Clique in the Divisibility Graph DP

    http://codeforces.com/contest/566/problem/F F. Clique in the Divisibility Graph time limit per test ...

  3. Codeforces 459E Pashmak and Graph(dp+贪婪)

    题目链接:Codeforces 459E Pashmak and Graph 题目大意:给定一张有向图,每条边有它的权值,要求选定一条路线,保证所经过的边权值严格递增,输出最长路径. 解题思路:将边依 ...

  4. Codeforces Round #261 (Div. 2) E. Pashmak and Graph DP

    http://codeforces.com/contest/459/problem/E 不明确的是我的代码为啥AC不了,我的是记录we[i]以i为结尾的点的最大权值得边,然后wa在第35  36组数据 ...

  5. Codeforces.566F.Clique in the Divisibility Graph(DP)

    题目链接 \(Description\) 给定集合\(S=\{a_1,a_2,\ldots,a_n\}\),集合中两点之间有边当且仅当\(a_i|a_j\)或\(a_j|a_i\). 求\(S\)最大 ...

  6. 63. Unique Paths II (Graph; DP)

    Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How m ...

  7. 62. Unique Paths (Graph; DP)

    A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The ...

  8. 64. Minimum Path Sum (Graph; DP)

    Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which ...

  9. codeforces 459E E. Pashmak and Graph(dp+sort)

    题目链接: E. Pashmak and Graph time limit per test 1 second memory limit per test 256 megabytes input st ...

随机推荐

  1. 学习笔记之Kubernetes

    Kubernetes | Production-Grade Container Orchestration https://kubernetes.io/ Kubernetes is an open-s ...

  2. 学习笔记之REST/RESTful

    REST(Representational state transfer) - Wikipedia https://en.wikipedia.org/wiki/Representational_sta ...

  3. selenium java-2 chrome driver与对应版本

    chrome driver下载地址:https://npm.taobao.org/mirrors/chromedriver driver与chrome的对应关系: 1.进入最新的driver,查看no ...

  4. BOM及改变this指向

    bom ( borwser object model 浏览器对象模型) 定义js操作浏览器的属性和方法 window.open(url way())    中有两个参数 url代表打开的网页地址 wa ...

  5. 高通QMI协议

    QMI(Qualcomm MSM Interface,官方名称应该是Qualcomm Message Interface)是高通用来替代OneRPC/DM的协议,用来与modem通信. QMI协议定义 ...

  6. python中for...if...构建List

    1.简单的for...[if]...语句 >>> a=[12, 3, 4, 6, 7, 13, 21] >>> newList = [x for x in a] & ...

  7. C# 获取物理网卡Mac地址

    // <summary> /// 获取网卡物理地址 /// </summary> /// <returns></returns> public stat ...

  8. ESB初步配置文件认识

    每个项目的都有各自的场景,但是其实往小处说,场景的处理基本都是很相似,之前做copy文件的程序,其实就是一种很常见的ETL的过程(转移文件,异构系统通过文件系统交换数据,存在数据同步). 了解一下ET ...

  9. php 编程笔记分享

    php获取POST数据的三种方法php 图片加水印源代码php+ajax+json的一个最简单实例php 汉字转拼音源码php遍历目录,生成目录下每个文件的md5值并写入到结果文件中php实现linu ...

  10. 模板方法模式( TemplateMethod)

    定义一个操作中的算法的骨架,而将一些步骤延迟到子类中,模板方法使得子类可以不改变一个算法的结构即可重新定义该算法的某些特定步骤. AbstractClass 是抽象类,其实也是一个抽象模板,定义并实现 ...