HDU 1024 Max Sum Plus Plus（dp）

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20107    Accepted Submission(s): 6638

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）

 

Recommend

/*
题意:n个数划分为m个集合，求集合最大值
思路:dp[i][j]代表 前i个数划分为j个集合的最大值

那么转移方程 dp[i][j]=max(dp[k][j-1],dp[i-1][j])+a[i]
                              j-1=<k<=i
这里的j能够滚动来节约空间

*/


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)


#define debug printf("%d
",bug++)
#define eps 1e-8
typedef __int64 ll;

using namespace std;


#define INF 0x3f3f3f3f
#define N  1000005

ll dp[N][2];

int bug;


int m,n;
int a[N];

int main()
{
    int i,j;
   // freopen("H:/in.txt","r",stdin);
    while(~scanf("%d%d",&m,&n))
    {
        bug=0;
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);

        for(i=0;i<=n;i++)
            dp[i][0]=dp[i][1]=-1111111111111;
        dp[0][0]=dp[0][1]=0;
        int cur=0;

        for(i=1;i<=m;i++)
        {
            dp[i][cur]=dp[i-1][cur^1]+a[i];

            ll ma=dp[i-1][cur^1];  //前i-1个数组合成i-1个集合的最大值

            for(j=i+1;j<=n;j++)
            {
               ma=max(dp[j-1][cur^1],ma);
               dp[j][cur]=max(ma,dp[j-1][cur])+a[j];
                        //前j-1个数组合成i-1个集合的最大值  前j-1个组合成i个集合加上这个数
            }
            cur^=1;
        }
       ll ans=-1111111111111;
       cur^=1;
       for(i=m;i<=n;i++)
         if(dp[i][cur]>ans) ans=dp[i][cur];
         printf("%I64d
",ans);
    }
    return 0;
}