杭电1024（Max Sum Plus Plus）

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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 ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + 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 S₁, S₂, S₃ ... 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

思路：

经典的动态规划优化的问题：设f(i, j)表示前i个数划分成j段，且包括第i个数的最大m子段和，那么有dp方程：

f(i, j) = max { f(i - 1, j) + v[i], max {f(k, j - 1) + v[i]}(k = j - 1 ... i - 1) } 。可以引入一个辅助数组来优化转移。设g(i, j)表示前i个数划分成j段的最大子段和（注意第i个数未必在j段里面），那么递推关系如下： g(i, j) = max{g(i - 1, j), f(i, j)}，  分是否加入第i个数来转移 这样f的递推关系就变成： f(i, j) = max{f(i - 1, j), g(i - 1, j - 1)}
+ v[i]，这样最后的结果就是g[n][m]，通过引入辅助数组巧妙的优化了转移。实现的时候可以用一维数组，速度很快 g[i][j]要么和g[i-1][j]相等，要么和f[i][j]相等 ，f[i][j]-a[i]要么和g[i-1][j-1]相等，要么和f[i-1][j]相等 。转成一维数组 到第i行时：f[j]=max{ f[j],g[j-1] }+a[i] } g[j]=max{ g[j],f[j] }。

代码实现：

import java.util.*;
class Main{
    public static void main(String[] args){
        int j;
        Scanner sc=new Scanner(System.in);
        while(sc.hasNext()){
            int m=sc.nextInt();int n=sc.nextInt();
            int[] a=new int[n+1];int[] dp=new int[m+1];int[] dp2=new int[m+1];
            for(int i=1;i<=n;i++){
                a[i]=sc.nextInt();
            }
            dp[1]=a[1];dp2[1]=a[1];
            for(int i=2;i<=n;i++){
                for(j=1;j<=Math.min(i, m);j++){
                    dp2[j]=Math.max(dp2[j]+a[i], dp[j-1]+a[i]);
                    dp[j-1]=Math.max(dp[j-1], dp2[j-1]);
                }
                dp[j-1]=Math.max(dp[j-1], dp2[j-1]);
            }
            System.out.println(dp[m]);
        }
    }
}