HDU-1024 java 实现

最新推荐文章于 2019-11-04 14:17:18 发布

SoulGuide

最新推荐文章于 2019-11-04 14:17:18 发布

阅读量1.1k

点赞数

CC 4.0 BY-SA版权

分类专栏： HDU-动态规划文章标签： HDU-Java

本文链接：https://blog.youkuaiyun.com/SoulGuide/article/details/79646233

HDU-动态规划专栏收录该内容

8 篇文章

订阅专栏

本文探讨了MaxSumPlusPlus问题，这是一个经典的算法问题，要求在给定的连续整数序列中找到m对下标，使得这m段序列的和达到最大值。通过动态规划方法解决，并提供了一个完整的Java实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33847 Accepted Submission(s): 12057

Problem Description

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S₁, S₂, S₃, S₄ ... S_x, ... S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767). We define a function sum(i, j) = S_i + ... + S_j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ j_x or i_x ≤ j_y ≤ j_x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i_x, j_x)(1 ≤ x ≤ m) instead. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1 2 32 6 -1 4 -2 3 -2 3

Sample Output

最大M子段和，求一个数组中分成M段取数，最大的和是多少。

import java.util.Scanner;
public class pro1024 {
	public static void main(String[] args) {
		// TODO 自动生成的方法存根
      Scanner cin=new Scanner(System.in);
      while(cin.hasNext())
      {
    	  int m=cin.nextInt();
    	  int n=cin.nextInt();
    	  int []num=new int[n+1];
    	  int dp[]=new int[n+1];
    	  for(int i=1;i<=n;i++)
    	  {  
    		  num[i]=cin.nextInt();
    	  }
    	 int mmmax=-2100000000;
    	  int jmax[]=new int [n+1];  
    	  for(int i=1;i<=m;i++)
    	  {
    		  mmmax=-2100000000;
    		  for(int j=i;j<=n;j++)
    		  {
    			  dp[j]=Math.max(dp[j-1]+num[j], jmax[j-1]+num[j]);			
                  jmax[j-1]=mmmax;
                  mmmax=Math.max(mmmax, dp[j]);
    		  }
    	  }
    	  System.out.println(mmmax);
      }
	}

}

核心代码来自于http://blog.youkuaiyun.com/youchengyuanzhi/article/details/8875354