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): 15817    Accepted Submission(s): 5140

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_xor 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）

 

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//609MS    8460K    692 B    C++
/*

    求最大m段不重叠子序列和。
    
    DP：
    状态转移方程：
        dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) 0<k<j
        dp[i][j]表示i段时遍历到j并使用a[j]时最大值
        
    使用滚动数组 maxn[] 记录 max(dp[i-1][k]) 

*/
#include<stdio.h>
#include<string.h>
#define N 1000005
#define inf 0x7fffffff
int dp[N],maxn[N],a[N];
inline int Max(int x,int y)
{
    return x>y?x:y;
}
int main(void)
{
    int n,m;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        int tmaxn;
        memset(dp,0,sizeof(dp));
        memset(maxn,0,sizeof(maxn));
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=m;i++){
            tmaxn=-inf;
            for(int j=i;j<=n;j++){
                dp[j]=Max(dp[j-1],maxn[j-1])+a[j];
                maxn[j-1]=tmaxn;
                tmaxn=Max(tmaxn,dp[j]);        
            }
        }
        printf("%d
",tmaxn);
    }
    return 0;
}