Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35988 Accepted Submission(s): 12807

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

Hint

Huge input, scanf and dynamic programming is recommended.

Author

JGShining（极光炫影）

分析：

m个子段的最大子段和

子段不重合

第一次写这个问题，还不是很懂

我是参考的大佬的博客：

https://www.cnblogs.com/dongsheng/archive/2013/05/28/3104629.html

回来更新一波

今天训练赛又遇到这个问题了，说说我的理解

这个问题的dp做其实跟LIS和LCS的有一丁点的像的

外层n循环，i

内层 0到i-1循环 j

每第i次选择都是在前面的i个中（不包括i）选最大的加上当前a【i】的值

上面都是自己瞎逼逼，不好理解，看下面

分析：

最大m段字段和问题是由最大字段和问题演变来的

最大字段和问题大家应该都知道

最大字段和是把每个数据看成一个对象，从而组成一个子段

核心：那么最大m段字段和，就是把一个子段看成一个对象，再进行最大字段和的操作，从而组成m个最大子段，所以是对象变了（自己也是突然想到）

code：

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

#define INF 0x7fffffff//无穷大
#define max_v 1000010
int a[max_v];
int now[max_v];// now[j]:包含第j个元素的最大和
int pre[max_v];// pre[j]:前j个元素的最大和，不包括第j个元素
int main()
{
    int n,m,maxx;
    while(~scanf("%d %d",&m,&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        memset(now,0,sizeof(now));
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=m;i++)//m个子段
        {
            maxx=-INF;
            for(int j=i;j<=n;j++)
            {
                now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);
                //now[j] 有两种来源，一种是直接在第i个子段后面加a[j]，一种是a[j]单独成为一个子段
                pre[j-1]=maxx;//更新pre使得pre[j-1]是前j-1个中的最大子段和
                maxx=max(now[j],maxx);
            }
        }
        printf("%d
",maxx);
    }
    return 0;
}