HDU 1024 Max Sum Plus Plus（m个子段的最大子段和）

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1024

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 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.

 

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

 

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define max_v 1000010
#define INF 0x7fffffff//无穷大
#define MIN_SUM 0x80000000//无穷小
int a[max_v];
int now[max_v];//now[j-1]表示的是以j-1结尾的元素i个子段的数和
int pre[max_v];//pre[j-1]表示的是前j-1个元素中i-1个子段的数和
int main()
{
    int n,m,maxx,i,j;
    while(~scanf("%d %d",&m,&n))
    {
        for( i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(now,0,sizeof(now));
        memset(pre,0,sizeof(pre));
        for(i=1;i<=m;i++)
        {
            maxx=-INF;
            for(j=i;j<=n;j++)
            {
                now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);
                pre[j-1]=maxx;//放在此处是为了实现pre[j-1]+a[j]中a[j]是一个独立的子段，那么此时就应该用的是i-1段
                if(now[j]>maxx)
                {
                    maxx=now[j];
                }
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}