hdu 1024 dp

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, S₂, 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 32 6 -1 4 -2 3 -2 3

Sample Output

68          

Hint

Huge input, scanf and dynamic programming is recommended.

题意：加强版的最大子段和，现在不是找一段了，是找不相交的m段的最大和

思路：dp[i][j]表示的是第j个数出现在第i段的最大值

状态转移方程：dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k]));

就是分为了两种情况

1.是第j个数作为第i段的第一个数

2.不是第一个数

优化：但是这个的数据量特别大，直接做的话既会超时又会超空间

空间优化：因为根据状态转移方程我们需要的最多也就两维，所以用滚动数组就可以了，在优化的话，其实直接开一维数组就可以

时间优化：我们可以优化的地方其实就是三重循环的最内层，因为我们其实不需要每次都循环计算，会有好多重复的计算，我们只需要维护一下前边的和就可以了

ac代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int a[1000005];int dp[1000005];int pre[1000005];int main(){    int n,m;    while(cin>>m>>n)    {       for(int i=1;i<=n;i++)       scanf("%d",&a[i]);       memset(dp,0,sizeof(dp));       memset(pre,0,sizeof(pre));       int max1;       for(int i=1;i<=m;i++)       {           max1=-100000000;           for(int j=i;j<=n;j++)           {              dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j]);              pre[j-1]=max1;              max1=max(max1,dp[j]);           }       }       cout<<max1<<endl;    }    return 0;}