hdu 1087 Super Jumping! Jumping! Jumping! 动态规划

Super Jumping! Jumping! Jumping!

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

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

   
   
   
   

 

这道题的实质就是求一个递增序列，使得组成该序列的所有元素的和最大，输出和即可。

解题思路：判断i点元素和前面元素能否构成递增序列，如果能，则求出该序列在i点的和；若不能，则把a[i]作为i点的和。最后求出最大的和即可。

AC代码：

#include<stdio.h>
int a[1005],sum[1005];
int main()
{
    int n,i,j;
    while(scanf("%d",&n)&&n)
    {
        for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        int max=sum[0]=a[0]; //定义a[0]为最大值
        for(i=1;i<n;i++)
         {
          sum[i]=a[i];// 先把它自己作为序列在i点的和
          for(j=0;j<i;j++)
            if(a[i]>a[j]&&sum[j]+a[i]>sum[i])
            //如果序列是递增的，且前面比它小的元素组成的递增序列的和加上它自己，大于它自己
              sum[i]=a[i]+sum[j];//把二者的和作为递增序列在i点的和
            if(sum[i]>max)  //如果递增序列在i点的和大于最大值
            max=sum[i]; //把在i点的和作为最大值
         }
        printf("%d\n",max);
    }
    return 0;
}