hdu 1087 Super Jumping! Jumping! Jumping!(动态规划DP)

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原文链接：http://www.cnblogs.com/GetcharZp/p/9054134.html

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#数据结构与算法 #java #c/c++

本文介绍了一种名为SuperJumping的棋盘游戏，玩家需从起点跳至终点，途中经过的棋子数值累加为得分。文章详细解析了使用动态规划求解最大得分的算法，包括状态方程和核心代码实现。

Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47017 Accepted Submission(s): 21736

Problem Description
Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.

The game can be played by two or more than two players. It consists of a chessboard（棋盘）and some chessmen（棋子）, and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
Your task is to output the maximum value according to the given chessmen list.

Input
Input contains multiple test cases. Each test case is described in a line as follow:
N value_1 value_2 …value_N
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
A test case starting with 0 terminates the input and this test case is not to be processed.

Output
For each case, print the maximum according to rules, and one line one case.

Sample Input
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0

Sample Output
4
10
3

分析：

　　①、动态规划（局部最优问题 ==> 全局最优）

　　②、状态方程：dp[i] = max(dp[i], dp[j] + A[i])

步骤：

　　①、从左到右依次遍历考虑该点的前面所有情况

　　②、通过状态方程 dp[i] = max(dp[i], dp[j] + A[i]) 计算该对应点的局部最优

　　③、通过局部最优 ==> 推出全局最优

核心代码：

1 for(int i = 0; i < n; ++ i)
2 {
3     dp[i] = A[i];
4     for(int j = 0; j < i; ++ j)
5     {
6         if(A[j] < A[i]) dp[i] = max(dp[i], dp[j] + A[i]);
7     }
8     my_max = max(my_max, dp[i]);
9 }

C/C++代码实现（AC）：

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <cmath>
 6 #include <stack>
 7 #include <map>
 8 #include <queue>
 9 
10 using namespace std;
11 const int MAXN = 1010;
12 long long A[MAXN], dp[MAXN], my_max;
13 
14 
15 int main()
16 {
17     int n;
18     while(~scanf("%d", &n), n)
19     {
20         memset(dp, 0, sizeof(dp));
21         my_max = -0x3f3f3f3f;
22         for(int i = 0; i < n; ++ i)
23             scanf("%d", &A[i]);
24 
25         for(int i = 0; i < n; ++ i)
26         {
27             dp[i] = A[i];
28             for(int j = 0; j < i; ++ j)
29             {
30                 if(A[j] < A[i]) dp[i] = max(dp[i], A[i] + dp[j]);
31             }
32             my_max = max(my_max, dp[i]);
33         }
34         printf("%lld\n", my_max);
35     }
36     return 0;
37 }

转载于:https://www.cnblogs.com/GetcharZp/p/9054134.html