Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39886 Accepted Submission(s): 14338

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...Sx,...Sn(1≤x≤n≤1,000,000,−32768≤Sx≤32767)S_1, S_2, S_3, S_4 ... S_x, ... S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767)S1​,S2​,S3​,S4​...Sx​,...Sn​(1≤x≤n≤1,000,000,−32768≤Sx​≤32767). We define a function sum(i,j)=Si+...+Sj(1≤i≤j≤n)sum(i, j) = S_i + ... + S_j (1 ≤ i ≤ j ≤ n)sum(i,j)=Si​+...+Sj​(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(im,jm)sum(i_1, j_1) + sum(i_2, j_2) + sum(i_3, j_3) + ... + sum(i_m, j_m)sum(i1​,j1​)+sum(i2​,j2​)+sum(i3​,j3​)+...+sum(im​,jm​) maximal (ix≤iy≤jxi_x ≤ i_y ≤ j_xix​≤iy​≤jx​ or ix≤jy≤jxi_x≤ j_y ≤ j_xix​≤jy​≤jx​ 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(ix,jx)(1≤x≤m)sum(i_x, j_x)(1 ≤ x ≤ m)sum(ix​,jx​)(1≤x≤m) instead.

Input

Each test case will begin with two integers $m$ and $n$, followed by $n$ integers S1,S2,S3...SnS_1, S_2, S_3 ... S_nS1​,S2​,S3​...Sn​.

Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1 2 32 6 -1 4 -2 3 -2 3

Sample Output

68

题意

将一个长度为$n$的数组分成不相交的$m$段，求这$m$段的和的最大值

思路

状态：$dp[i][j]$表示在前$j$个数中取出$i$段的最大和

状态转移方程：$dp[i][j]=max(dp[i-1][k],dp[i][j-1])+num[j](i-1\le k\le j-1)$

由于$m$范围未知，n≤106n\leq 10^6n≤106，所以二维的dp方程无论是在时间上还是在空间上都是不允许的。

那么我们就需要对这个方程进行优化：

不难发现当前状态只与两个状态有关：

1. 第$j$个数和前$j-1$个数在一段里
2. 第$j$个数和前$j-1$个数不在一段里。

根据这一点，我们把状态降成一维的数组，$dp[j]$表示前$j$个数分$i$段时的最大和，然后用$sum[j-1]$来表示状态一的前$j-1$个数在前$i-1$段的最大和，$dp[j-1]$表示状态二的前$j-1$个数在前$i$段的最大和。

当前状态的转移方程为：$dp[j]=max(dp[j-1],sum[j-1])+num[j]$，持续更新dp与sum数组的值

AC代码

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