Hdu1087-Super Jumping! Jumping! Jumping!-【dp动态规划】

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本文介绍了一款名为“SuperJumping!Jumping!Jumping!”的游戏算法实现，重点在于如何通过动态规划寻找棋盘上从起点到终点的最大得分路径，并提供了一个具体的编程实现案例。

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传送门：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): 40462 Accepted Submission(s): 18694

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

 
 状态转移公式：dp[i]=max{dp[j]+num[i]}(num[i]>num[j],j<i);

i代表以第i个数结尾的最大上升子序列的和。
题目思路：dp数组表示是包含当前这个数的最大递增子序列和。dp[i]表示的是前i个并且包含第i个的最大递增子序列和！给个数据：3 1 4 显然dp[1]=3,dp[2]=1表示两个数的最大值。因为分两种情况讨论，如果第二个数大于第一个数，就加上，即dp[2]=dp[1]+num[2]；如果不大，dp[2]=num[2];dp[3]=7表示三个数的最大值。首先比较num[3]和num[1],如果num[3]>num[1],dp[3]=7先存下来,如果num[3]>num[2],dp[3]=5依旧存下来;还有一种如果num[3]比前两个值都小，dp[3]=num[3];最后在存下来的dp[3]中找到一个最大的！

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
#define INF 0x3f3f3f3f
int dp[1050],a[1050];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF&&n)
	{
		for(int i=0; i<n; i++)
		{
			scanf("%d",&a[i]);
			dp[i]=a[i];
		}
		int sum=-INF;
		for(int i=0; i<n; i++)
		{
			for(int j=0; j<i; j++)
			{
				if(a[i]>a[j])
				dp[i]=max(dp[i],dp[j]+a[i]);
			}
			sum=max(sum,dp[i]);
		}
		printf("%d\n",sum);
	}
	return 0;
}
/*
Sample Input
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0
Sample Output
4
10
3
*/