hdu1024Max Sum Plus Plus【状态dp 滚动数组】

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 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, S ₂, 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 3 2 6 -1 4 -2 3 -2 3

 

Sample Output

6 8

1003的升级版==就像之前看1505city game 和 1506Largest Rectangle in a Histogram 没看出来二者有什么关系一样 ==

这次做完了1003maxsum 依旧没做出来1024更何况1003的写法就不对→_→

状态转移方程 dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) 前者是与之前的成为一组 后者是独立成组

dp[i][j]表示前j个数组成i组 而dp[i-1][k]是在j=j-1时的mmmax （怪不得他的序号就是j-1==）即mmax[j-1]

而mmax[]只有在下一次i=i+1 j=j 这个对应位置会被用到 恰恰说明了dp[i-1][k] 的本质是为了取上一次循环中的最大值

#include 
#include
#include
using namespace std;
int a[1000005],dp[1000005],mmax[1000005],n,m,mmmax;
int main()
{
    while(~scanf("%d%d",&m,&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        memset(mmax,0,sizeof(mmax));
        for(int i=1;i<=m;i++)
        {
            mmmax=-0x3f3f3f3f;
            for(int j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1]+a[j],mmax[j-1]+a[j]);
                mmax[j-1]=mmmax;
                mmmax=max(dp[j],mmmax);
            }
        }
        printf("%d\n",mmmax);
    }
    return 0;
}