hdu 1024

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

这个题的意思是找m个不相交的区间，求他们的最大和，这个题，如果我们用二维的数组进行dp，dp[i][j]=max(dp[i][j-1]+num[j],dp(i-1,t)+num[j])   其中i-1<=t<=j-1，其中，dp[i][j]表示以a[j]结尾而有i个间的最大和，但是这就提高了空间复杂度，怎么办呢，注意，这里状态转移的时候既有dp[i]的数组，又有dp[i-1]的数组，这导致我们dp的时候不能像背包一样只开一个一维数组，因为后面的状态必定要覆盖掉前面的状态，这样我们只能另外找一个数组，把i-1的状态存起来，这个和做划分数的一维数组做法是一模一样的！！注意状态转移的顺序，pre表示i-1的状态，为什么之前背包可以用一维数组做，就是因为状态转移的顺序决定的！！！

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[1000004];
int dp[1000004];
int pre[1000004];
int main()
{
    int n,m;
    while(~scanf("%d%d",&m,&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            dp[i]=pre[i]=0;
        }
        int res=0;
        dp[0]=pre[0]=0;
        for(int i=1;i<=m;i++)
        {
            res=-100000000;
            for(int j=i;j<=n;j++)
            {
                 dp[j]=max(dp[j-1],pre[j-1])+a[j];
                 pre[j-1]=res;
                 res=max(dp[j],res);
            }
        }
        cout<<res<<endl;
    }
}