HDOJ 1024 Max Sum Plus Plus

dp[ i ][ j ]表示从1～～ j  分了 i 段，并且以 a[ j ]结尾的最大长度。

dp[ i ][ j ] = max( dp[ i ][ j-1] ,  dp[ i-1 ] [ k ]) +a[ j ]     即a[ j ] 可以接在最后一段后面，也可以另起一段

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12622    Accepted Submission(s): 4159

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）

 

 

 #include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

int a[1000100];

int dp[2][1000100];

int sum[1000100];

int f[1000100];

int n,m;

int main()

{

while(scanf("%d%d",&m,&n)!=EOF)

{

    memset(dp,0,sizeof(dp));

    sum[0]=0;

    for(int i=1;i<=n;i++)

    {

        scanf("%d",&a);

        sum=sum[i-1]+a;

    }

    int ans=-99999999;

    int cur=0;

    for(int j=1;j<=m;j++)

    {

        for(int i=0;i<=j-1;i++)

            dp[1-cur]=-999999999;

        for(int i=j;i<=n;i++)

        {

            if(i==j)  f=sum;

            else f=max(dp[cur][i-1],f[i-1])+a;//如果用3重循环就超时了

            dp[1-cur]=max(dp[1-cur][i-1],f);

            if(j==m)

                ans=max(ans,f);

        }

        cur=1-cur;

    }

    printf("%d\n",ans);

}

    return 0;

}