Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23976 Accepted Submission(s):
8199

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
const int MAX = 1e6 + 10;
const int MIN = -1e8;

int curr[MAX], pre[MAX], a[MAX];
int max_sum, n, m;

int main()
{
    while(cin >> m >> n)
    {
        for(int i = 1; i <= n; i++)
            cin >> a[i];
        memset(curr,0,sizeof(curr));
        memset(pre,0,sizeof(pre));
        int j = 0;
        for(int i = 1; i <= m; i++)
        {
            max_sum = MIN;
            for(j = i; j <= n; j++)
            {

                curr[j] = max(curr[j - 1], pre[j - 1]) + a[j];
/*
    pre[j-1]表示的是上一个状态中i...j-1的最大值，
    max_sum更新之后表示的i...j的最大值，所以不能写反了
*/
                pre[j - 1] = max_sum;
                max_sum = max(max_sum, curr[j]);
            }
            //pre[j-1]中始终保持着前一个状态的最大值，这个很重要
            pre[j - 1] = max_sum;
        }
        cout << max_sum << endl;
    }
    return 0;
}