Max Sum Plus Plus HDU - 1024

    Max Sum Plus Plus     HDU - 1024

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

Output

Output the maximal summation described above in one line. 

and n, followed by n integers S ₁, S₂, S ₃ ... S _n. 
Process to the end of file.

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.

 

分析：d[j]为遍历到当前数字时，前j-1个数的i段区间和的最大值加上a[j]，d[j]是动态的，可取可不取；

p[j]为记录前j个数的i段区间和的最大值；

 

	j	1	2	3	4	5	6
i		-1	4	-2	3	-2	3
1	d	-1	4	2	5	3	6
1	p	-1	4	4	5	5
2	d	-1	3	2	7	5	8
2	p	-1e9	3	2	7	7

代码中p[n]没有记录，直接放在了ans中。

代码：

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<math.h>

using namespace std;

const int Max=1e6+5;

int p[Max],a[Max],d[Max];

int main()

{

     int m,n,ans;

     while(scanf("%d%d",&m,&n)!=EOF)

     {

          memset(p,0,sizeof(p));

          memset(d,0,sizeof(d));

          for(int i=1;i<=n;i++)

          scanf("%d",&a[i]);

          for(int i=1;i<=m;i++)

{

               ans=-1e9;

               for(int j=i;j<=n;j++)

               {

                    d[j]=max(d[j-1],p[j-1])+a[j];

                    p[j-1]=ans;

                    ans=max(ans,d[j]);

               }

          }

          printf("%d\n",ans);

     }

}