"max m and" dp + scrolling arrays

Title Description

Given the number of n, the number of n is divided into the maximum m sub-segments of the M-segment Disjoint.

Gives an integer sequence of a1,a2,a3,a4,...,ax,...,an, wherein 1≤x≤n≤1,000,000, -32768≤sx≤32767.

We define a function sum (i,j) =ai + ... + Aj (1≤i≤j≤n, and Ai~aj is a sequential number).

Now that we get a positive integer m (1≤x≤m≤30), your job is to look for M to I with J. This m meets the following conditions for I and J:
Sum (i1,j1) +sum (i2,j2) +sum (i3,j3) +...+sum (im,jm) is the largest in this sequence.
Note: the intersection of any two intervals [ix,jx] and [iy,jy] is an empty set.

Please find out: what is the maximum sum (i1,j1) +sum (i2,j2) +sum (i3,j3) +...+sum (im,jm)?

Input Format

The first row of two integers n,m.
The second line is an n integer separated by a space, that is, A1~an.

output Format

Please output the maximum value of our defined: sum (i1,j1) +sum (i2,j2) +sum (i3,j3) +...+sum (im,jm), which is the largest m sub-segment and.

Sample Data 1input

3 1
1 2 3

Output

6

Sample Data 2input

6 2
-1 4-2 3-2 3

Output

8

Problem Analysis

The problem of "f[j][i x, divided into y-parts", usually DP is a "," which means dividing the first J elements into the most * values of the I Parts.

The problem is the same: f[j][i] as mentioned above, there are two decisions for newly added elements:

1. Add to the last paragraph of the previous One.
2. Separate into Segments.

The transfer equation is: $f [j][i] = max (f[j-1][i] + val[i], max_{k = 1}^{j-1}\{f[j-1][k]\} + val[i]) $

A little calculation, you will find that such a space has been report. The next step is the scrolling array.

It can be found that the J dimension of the transfer equation is always shifted from the last j-1, and I is not changed, and f[j-1][k] is only required to be the maximum value of the last time, then the following optimizations can be made using a scrolling array:

1. Remove the I this dimension, and I this one dimension refers to the outer cycle, the guarantee j-1 to J is I this one dimension. Array requires only f[n]

2. Use mx[j-1] to represent $max_{k = 1}^{j-1}\{f[j-1][k]\} $, with each internal loop (j) update for the next outer loop (i).

So the shift becomes:

 for (int1; I <= m; i++) {        =-oo;          for (int j = i; J <= n; J + +) {            11] + val[j]            ); 1 ]);             1] = tmp;        }    }

Code

#include <iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<string>#include<cmath>using namespacestd;Const intN = 1e6 +5, oo =0x7fffffff;intn, m;typedefLong Longll;ll f[n], mx[n], val[n];inline ll read () {ll i=0, F =1;CharCH =GetChar ();  for(; (ch <'0'|| CH >'9') && ch! ='-'; CH =GetChar ()); if(ch = ='-') F =-1, ch =GetChar ();  for(; CH >='0'&& CH <='9'; CH =GetChar ()) i= (i <<3) + (i <<1) + (ch-'0'); returnIf;} InlinevoidWR (ll X) {if(x <0) Putchar ('-'), x =-x; if(x >9) WR (x/Ten); Putchar (x%Ten+'0');}intmain () {n= Read (), m =Read ();  for(inti =1; I <= n; I++) val[i] =Read ();  for(inti =1; I <= m; i++) {ll tmp= -oo;  for(intj = i; J <= n; J + +) {f[j]= Max (f[j-1] + val[j], mx[j-1] +val[j]); TMP= Max (tmp, f[j-1]); Mx[j-1] =tmp; }} ll ans= -oo;  for(inti = m; I <= n; i++) ans=Max (ans, f[i]); WR (ans), Putchar ('\ n'); return 0;}

"max m and" dp + scrolling arrays