hdu 1024 Max Sum Plus Plus(DP最大字段和)

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本文探讨了MaxSumPlusPlus问题，这是一种比经典MaxSum更复杂的动态规划问题。任务是在给定序列中找到m对子序列，使得这些子序列的和达到最大值。文章提供了详细的算法思路和实现代码。

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11158 Accepted Submission(s): 3673

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）

思路：s[i][j]为i段包含j的最大和。ma[i][j]为前j个分为i段的最大和。答案就是ma[m][n]

d[j]存的是数

s[i][j]=max(s[i][j-1]+d[j],ma[i-1][j-1]+d[j]);

ma[i][j]=max(s[i][k]) (i<=k<=j)

又因为题目数给的太大一百万。所以二维数组不好开。想想可以用一维。

s[j]为i段含j最大和，ma【j】为i段前j个最大和。

s[j]=max(s[j-1]+d[j],ma[j-1]+d[j])

ma[j]=max(s[k]) （i<=k<=j）

ma[j]要先用后更新。这样就不会有问题了。

#include<cstdio>
#include<cstring>
const int mm=1000010;
const int oo=1e9;
int d[mm],s[mm],ma[mm];
int n,m;
int main()
{
  while(scanf("%d%d",&m,&n)!=EOF)
  {
    for(int i=1;i<=n;i++)
      scanf("%d",&d[i]);
    memset(s,0,sizeof(s));
    memset(ma,0,sizeof(ma));
    int _min,_max;
    for(int i=1;i<=m;i++)
    { _max=-oo;int j;
      for(j=i;j<=n;j++)
       {if(s[j-1]<ma[j-1])
        s[j]=ma[j-1]+d[j];
        else s[j]=s[j-1]+d[j];
        ma[j-1]=_max;
        if(s[j]>_max)
          _max=s[j];
       }
       ma[j-1]=_max;
    }
    printf("%d\n",_max);
  }
}