HDU 1087 Super Jumping! Jumping! Jumping!（DP，上升子序列）

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1087

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32391    Accepted Submission(s): 14601

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
  

 

题意：给你一串数，从起点（不在这N个点内）到终点（不在N个点内），点数 的上升路径。起点保证是最小的数，终点保证是最大的数。就是保证一定可以从起点去任何位置，从任何位置到终点。

解题思路：思考相似的类型，最长上升子序列。设置一个dp数组，dp[i]代表以a[i]为结尾的上升子序列时，能够得到的最高分数。

我们可以考虑，以a[i]为结尾的子序列

①只包含a[i]的子序列

②在满足j < i 并且a[j] < a[i]的以a[j]为结尾的上升子序列，追加上a[i]后的子序列

我们就可以很容易得出递推式，dp[i] = max{ a[i] , dp[j] + a[i] | j < i 且 a[j] < a[i] };

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#include<cmath>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 1005;
int a[maxn];
int dp[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        for(int i = 0 ; i < n ; i ++)
            scanf("%d",&a[i]);
        int mmax  = 0;
        for(int i = 0 ; i < n ; i ++)
        {
            dp[i] = a[i] ;
            for(int j = 0 ; j < i ; j ++)
           {
               if(a[i]>a[j])
                dp[i] = max(dp[i],dp[j]+a[i]);
           }
           mmax = max(dp[i],mmax);
        }
        cout << mmax <<endl;
    }

    return 0;
}