HDU 1087Super Jumping! Jumping! Jumping! —简单DP

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本文介绍了一款名为“SuperJumping!Jumping!Jumping!”的游戏，通过动态规划求解数字串中最大递增子序列的问题。玩家需从起点跳跃至终点，途中访问的数字需递增，最终得分由路径上的数字总和决定。

原题链接：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): 21960 Accepted Submission(s): 9631

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

Sample Output

4
10
3

Author

lcy

题意：求数字串中数字和最大的递增子序列。

思路：简单DP，对于第i位数，dp[i]=max(dp[i],dp[j]+a[i])，a[i]为第i位数的值，j为在i之前所有比a[i]小的数一一列举。

AC代码：

#include "iostream"
using namespace std;

int dp[1001],ans,n,a[1001];

int solve()
{
	ans=0;
	memset(dp,0,sizeof(dp));
	for(int i=0;i=0;j--)
		{
			if(a[j]