hdu 1024



Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24332    Accepted Submission(s): 8356

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.

   
   
   
    
  
  

 

题意：

给出有n个元素的序列，将其分为m个字段，求字段和的最大值。（字段之间不能相交）。

dp[i][j]是前j个数够成i个字段的最大值。dp[i]要么与前面的数构成i个字段，要么独立的成为一个字段。方程：dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j])

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
#define MAXN 1000000
#define INF 0x3f3f3f3f
int dp[MAXN+10];
int pre[MAXN+10];
int a[MAXN+10];
int main()
{
    int n,m;
    int i,j,MAX;
    while(scanf("%d%d",&m,&n)!=EOF)  //m为分的字段数目，n序列中元素的数目 
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            pre[i]=0;
            dp[i]=0;
        }
        dp[0]=0; pre[0]=0;    
        for(i=1;i<=m;i++)
        {
            MAX=-INF;
            for(j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j]); // 第j个元素与前面元素组成i个子段或者第j个元素单独成一个字段 
                pre[j-1]=MAX;  //更新是pre[j-1]是前j-1个元素最大的子段和 
                if(dp[j] > MAX)
					MAX=dp[j];
            }    
        }  
        printf("%d\n",MAX);         
    } 
    return 0;   
}