HDU1024 Max Sum Plus Plus （优化线性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. ^_^

InputEach test case will begin with two integers m and n, followed by n integers S
₁, S
₂, S
₃ ... S
_n.

Process to the end of file.

OutputOutput 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.
别的博客在原理上说得很清楚了，代码里的注释解释了一下是如何进行时空优化的。

#include <bits/stdc++.h>
#define INF 1<<30
using namespace std;
int a[];
int dp[]={};//dp[i,j]表示选取a[j]分成i段的最大子段和 dp[i][j]=max(dp[i][j-1]+a[j],dp[i-1][k]+a[j])(k:i-1~j-1)
int m,n;
int pre[]={};
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(dp,,sizeof(dp));
        memset(pre,,sizeof(pre));
        int i,j;
        for(i=;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        int ans=;
        for(i=;i<=m;i++)
        {
            ans=-INF;//因为所求的最大状态一定在i=m这一轮出现，所以每轮开始把ans设为负无穷没有影响
            for(j=i;j<=n;j++)//从i开始，要不然的话少于i个数分不成i段了
            {
                dp[j]=max(dp[j-]+a[j],pre[j-]+a[j]);//在这里，一开始整个pre数组存储的本身就是i-1阶段里的各个最大值
                pre[j-]=ans;//当第二重循环跑起来时 ，在更新dp数组的同时也在更新pre数组，这就是这一句的含义，而在这一重循环中更新pre数组对当前是没有影响的（因为已经越过去了）
                ans=max(ans,dp[j]);//在开头初始化后，ans能保证一直是当前层的最大值
            }
        }
        cout<<ans<<endl;
    }
    return ;
 }