Super Jumping! Jumping! Jumping!（简单的动态规划） （F-6）

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53174 Accepted Submission(s): 24652

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

题目分析一下：
题意就是讲我们从start开始走棋子，下一步走到的地方数字要比上一步的大，可以跨越棋子来走。拿例子来讲，第一个数字n是指有多少个格子可以走，后面的n个数字分别是讲这n个格子所包含的数字，我们，目的就是求出这n个数的最大递增子序列的和（注意不一定是连续的），那么就简化为求n个数的最大递增子序列的和的问题了。
VJ通过的代码如下：

#include 
#include 
#define max(x,y) (x>y)?x:y
#define min(x,y) (x>y)?y:x
using namespace std;
int dp[2001], num[2001];//用num[]来存输入的n个数
int result;
int main()
{
	int n;
	while (scanf_s("%d", &n) != EOF&&n)
	{
		result = 0;
		for (int i = 0;i < n;i++)
			scanf_s("%d", &num[i]);
		memset(dp, 0, sizeof(dp));
		dp[0] = num[0];
		for (int i = 1;i < n;i++)
		{
			for (int j = 0;j < i;j++)
			{
				if (num[i] > num[j])//如果后面的一个数比前面的一个数大
				{
					dp[i] = max(dp[i], num[i] + dp[j]);//那么就加起来
				}
			}
			dp[i] = max(dp[i], num[i]);
		}
		for (int i = 0;i < n;i++)
		{
			result = max(result, dp[i]);
		}
		printf("%d\n", result);
	}
	return 0;
}