hdu 1024 dp

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：加强版的最大子段和，现在不是找一段了，是找不相交的m段的最大和

思路：dp[i][j]表示的是第j个数出现在第i段的最大值

状态转移方程：dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k]));

就是分为了两种情况

1.是第j个数作为第i段的第一个数

2.不是第一个数

优化：但是这个的数据量特别大，直接做的话既会超时又会超空间

空间优化：因为根据状态转移方程我们需要的最多也就两维，所以用滚动数组就可以了，在优化的话，其实直接开一维数组就可以

时间优化：我们可以优化的地方其实就是三重循环的最内层，因为我们其实不需要每次都循环计算，会有好多重复的计算，我们只需要维护一下前边的和就可以了

ac代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[1000005];
int dp[1000005];
int pre[1000005];
int main()
{
    int n,m;
    while(cin>>m>>n)
    {
       for(int i=1;i<=n;i++)
       scanf("%d",&a[i]);
       memset(dp,0,sizeof(dp));
       memset(pre,0,sizeof(pre));
       int max1;
       for(int i=1;i<=m;i++)
       {
           max1=-100000000;
           for(int j=i;j<=n;j++)
           {
              dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j]);
              pre[j-1]=max1;
              max1=max(max1,dp[j]);
           }
       }
       cout<<max1<<endl;
    }
    return 0;
}