hdu 1087Super Jumping! Jumping! Jumping!（简单的dp就是一个判断条件两个for循环，将这个量与前面出现的两进行比较）

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 22793    Accepted Submission(s): 10028

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

 

Author

lcy

题意：

进行从最小值找到最大值，求中间所经过的路径上的和的最大值。

难点：

如何判定此时的要经过的节点是最大值，需要遍历，与前面的节点的值进行比较

代码如下：

#include<stdio.h>int a[1010],s[1010];int main(){int n,i,j,maxi;while(~scanf("%d",&n),n){for(i=0;i<n;i++){scanf("%d",&a[i]);s[i]=a[i];}maxi=a[0];for(i=1;i<n;i++){for(j=i-1;j>=0;j--){if(a[j]<a[i]&&s[i]<s[j]+a[i])//题上的条件，后者大于前者，所以有前面的判断条件，s[i]在这里是暂时的存储总和的空间 s[i]=s[j]+a[i];if(s[i]>maxi)maxi=s[i];}}printf("%d\n",maxi);}return 0;}