HDU 题目1087 Super Jumping! Jumping! Jumping!

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21824    Accepted Submission(s): 9560

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 24 1 2 3 44 3 3 2 10

 

Sample Output

4103
这道题看不懂题。。。。是看了别人的翻译之后才知道的。
最大升序子序列的和
<pre name="code" class="html">#include <stdio.h>int dp[1005],a[1005];int main (void){  //  freopen("1087.txt","r",stdin);    int n,i,j,max;    while(scanf("%d",&n)!=EOF&&n)    {        max=-1;        memset(a,0,sizeof(a));        for(i=1;i<=n;i++)            scanf("%d",&a[i]);            memset(dp,-1,sizeof(dp));            dp[1]=a[1];        for(i=1;i<=n;i++)//动态规划的宗旨就是从最后往前来考虑，j的位置在i的位置的前边        {            dp[i]=a[i];//从每一个位置开始往前都当成一堆，然后将每一堆的最后一个位置初始化                for(j=1;j<=i;j++)//这是比较简单的动态规划            {                if(a[j]<a[i]&&dp[i]<dp[j]+a[i])                   dp[i]=dp[j]+a[i];            }            if(max<dp[i])//最后将各堆中确定的最大值互相比较，找出最大值，输出                max=dp[i];        }        printf("%d\n",max);    }    return 0;}