HDU1024 Max Sum Plus Plus —— DP + 滚动数组

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=1024

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31798    Accepted Submission(s): 11278

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

Hint

Huge input, scanf and dynamic programming is recommended.

 

Author

JGShining（极光炫影）

 

 

 

题解：

1.设last_max[i][j]为：处理到第j个数时，第j个数属于第i个序列的最大值。last_max[i][j]必定包含a[j]。

2.设all_max[i][j]为 ：处理到第j个数时， 分成i个序列的最大值。all_max[i][j]不一定包含a[j]， 其值实际上是 max( last_max[i][k] )   其中i<=k<=j，all_max数组的实际作用是：当a[j]独立出来自己作为一个序列时，找到前面i-1个序列和的最大值， 以便与a[j]拼接成i个序列。

3.状态转移：last_max[i][j] = max( last_max[i][j-1],  all_max[i-1][j-1] ) + a[j]。即以a[j]是否独立出来自己作为一个序列来讨论。

4.观察状态转移方程， last_max数组不需要知道上一步的信息，所以只需要开一维；all_max数组只需要知道上一步的信息，所以可以用滚动数组以节省空间。

 

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e6+10;

int a[MAXN];
int last_max[MAXN], all_max[2][MAXN];
int main()
{
    int n, m;
    while(scanf("%d%d",&m, &n)!=EOF)
    {
        for(int i = 1; i<=n; i++)
            scanf("%d", &a[i]);

        ms(all_max, 0);
        ms(last_max, 0);

        for(int i = 1; i<=m; i++)
        {
            last_max[i] = all_max[i&1][i] = last_max[i-1]+a[i];
            for(int j = i+1; j<=n; j++)
            {
                last_max[j] = max(last_max[j-1], all_max[!(i&1)][j-1]) + a[j];
                all_max[i&1][j] = max(all_max[i&1][j-1], last_max[j]);
                /**以上代码的实际意义为：
                last_max[i][j] = max(last_max[i][j-1], all_max[i-1][j-1]) + a[j];
                all_max[i][j] = max(all_max[i][j-1], last_max[i][j]);
                **/
            }
        }
        printf("%d
", all_max[m&1][n]);
    }
    return 0;
}