Algorithm _hdoj_1005

Question:

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 56626    Accepted Submission(s): 26273

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

 

Author

lcy

 

idea:

first create two arrays to saving the input data,one is data that use to saving the input data and it won't change anyway,the other one is result that storing the input data and will change every circle,it will store the sum of a increasing sub-sequence before this kont in the last, which every element is smaller than it.

it's a simply DP

 

translate in chinese:

Definition of two arrays, a first array is Data, used to store information entered, he will not change, and may only be accessed 
Further Result is an array, for storing a first node to each node increasing subsequence of and, finally this is the first storage array to increment a child node of the specified node sequence and 
then by traversing the result array can get the maximum increasing subsequence

View Code

 

source code:

import java.util.Scanner;

/**
 * 3 1 3 2
 * 4 1 2 3 4
 * 4 3 3 2 1
 * 0
 */
public class Main {
    static Scanner sc = new Scanner(System.in);
    public static void main(String[] args) {
       while(true){
           int ammount = sc.nextInt();
           if(ammount==0)break;
           int[] data = new int[ammount];
           int[] result = new int[ammount of]; 

           for ( int I = 0; I <ammount of; I ++ ) { 
               data [I] = sc.nextInt (); 
               Result [I] = data [I]; 
           } 

           / ** 
            * Thought: From the data array each of the first view, the first item from the beginning, if this is smaller than the specified item, to add, and then proceed to the next 
            * / 
           for ( int I =. 1; I <data.length; I ++ ) {
                for ( int J = 0; J <I; J ++ ) {
                    IF (Data [J] <Data [I]) Result [I] = Math.max (Result [I], Result [J] + Data [I]) ; 
               } 
           } 
           int ANS = -99;
           for(int temp:result){
               if(temp > ans)ans = temp;
           }
           System.out.println(ans);
       }
    }
}

 

hope that I can help you 

that's all