HDU - 1024 Max Sum Plus Plus （滚动数组）

最新推荐文章于 2024-07-21 14:30:00 发布

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最新推荐文章于 2024-07-21 14:30:00 发布

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分类专栏： dp

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本文探讨了MaxSumPlusPlus问题的解决方案，该问题是HDU ACM竞赛中一道经典的动态规划题目。任务是在给定的数列中找出m对下标(i, j)，使得这些下标对应的数列和总和最大，同时确保每对下标不重叠。文章提供了一个高效的算法实现，并通过样例输入输出展示了如何求解。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1024点击打开链接

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31448 Accepted Submission(s): 11114

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 S ₁, S ₂, S ₃, S ₄ ... 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(i ₁, j ₁) + sum(i ₂, j ₂) + sum(i ₃, j ₃) + ... + 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 S ₁, S ₂, S ₃ ... 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.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
using namespace std;
#define maxn 1000100
int dp[maxn];
int a[maxn];
int maxx[maxn];
int main()
{
    int n,m;
    while(~scanf("%d%d",&m,&n))
    {
        int temp=0;
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        memset(maxx,0,sizeof(maxx));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=m;i++)
        {
            temp=INT_MIN;
            for(int j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1],maxx[j-1])+a[j];
                maxx[j-1]=temp;
                temp=max(dp[j],temp);
            }
        }
        printf("%d\n",temp);
    }
}