HDU 1024 MaxSumPlusPlus（进阶一维dp）-优快云博客

本文介绍了一个基于动态规划的优化算法解决 MaxSumPlusPlus 问题的方法，该问题要求找出序列中 m 对下标，使得这些下标定义的子序列之和最大。通过对原始二维动态规划算法的空间优化，将其降至一维，从而有效解决了大规模数据输入下的内存限制问题。

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Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27506 Accepted Submission(s): 9586

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

Sample Output

6
8

Hint
Huge input, scanf and dynamic programming is recommended.

Author

JGShining（极光炫影）

题解：这道题是HDU1003的进阶版，当m=1时即HDU1003原题，通过观察题目我们很容易想到维护两个量的二维dp公式dp[i][j]表示考虑到前j位，划分i段的最大值，转移方程为dp[i][j]=max(dp[i][j-1],max(dp[i-1][k])(i<k<j))，但是再看一下空间限制和数据范围，这个二维数组要开到1e12，显然是会MLE的，因此，我们需要优化一下，也就是滚动更新dp数组，将这个二维数组降到1维。节省空间，即只需要两个一维数组维护dp[i][j-1]和max(dp[i-1][k])j即可；

代码如下

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

int a[1000005];
int d[1000005];
int p[1000005];
int mi=-1000000000;


int main()
{
    int n,m;
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        int mx;
        memset(d,0,sizeof(d));
        memset(p,0,sizeof(p));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=m;i++)
        {
            mx=mi;
            int j;
            for(j=i;j<=n;j++)
            {
                d[j]=max(d[j-1],p[j-1])+a[j];
                p[j-1]=mx;
                mx=max(mx,d[j]);
            }
            p[j-1]=mx;
        }
        cout<<mx<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}