hdu 1024 Max Sum Plus Plus(DP最大字段和)

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 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.

   
   
   
    
  
  

 

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);
  }
}