【基础dp】Max Sum Plus Plus

最新推荐文章于 2020-10-20 23:32:14 发布

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最新推荐文章于 2020-10-20 23:32:14 发布

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本文介绍了一道名为 MaxSumPlusPlus 的算法题目，通过动态规划方法求解给定序列中 m 对子序列的最大和。利用滚动数组优化内存使用，避免超时与超内存问题。

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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): 20745 Accepted Submission(s): 6900

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）

用正常的方法写会超内存超时间，所以此处用滚动数组；

进行m次操作，所以最外层循环为m次；

dp[i][j]表示处理到第i点，分成cnt组的总和，j表示滚动数组中的哪一个。

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;



int n,m;
int a[1000007];
int dp[1000007][2];

int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);

        int t1=0,t2=1;
        memset(dp,0,sizeof dp);

        for(int i=1;i<=m;i++)
        {
            dp[i][t2]=(dp[i-1][t1])+a[i];

            int ma=dp[i-1][t1];

            for(int j=i+1;j<=n;j++)
            {
                ma=max(ma,dp[j-1][t1]);
                dp[j][t2]=max(dp[j-1][t2],ma)+a[j];

            }
            swap(t1,t2);

        }

        int ans=-99999999;
        for(int i=m;i<=n;i++)
        {
            ans=max(ans,dp[i][t1]);
        }
        printf("%d\n",ans);

    }

    return 0;
}