hdu 1024 Max Sum Plus Plus (dp+二维变一维)

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本文介绍了一个挑战性的算法问题，即在给定连续数列的基础上，寻找特定组合的最大和。通过深入分析和优化DP（动态规划）方法，作者提出了一种高效的解决方案，并详细阐述了其实现过程。此外，文章还提供了输入输出样例，帮助读者理解问题的具体应用。

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 13722 Accepted Submission(s): 4522

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.

思路：

dp。

状态：dp[i][j]-最大和 i-已经组成多少组 j-已经用掉多少数。

方程：dp[i][j]=max(dp[i-1][j]+s[i],dp[k][j-1]+s[i]); (j>=i) (0<k<i)

时间复杂度为O(n*m^2),dp[k][j-1]可以递推时用一个值维护，降到O(n*m).

由于n比较大，空间吃不消，所以要优化空间。

考虑到第一维每次都是从i-1到i，故可以将第一维去掉。

方程变为 dp[j]=max(dp[j-1]+s[j],ma[j-1]+s[j]);

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 1000005
#define INF 0x3f3f3f3f
using namespace std;

typedef long long ll;
ll n,m,ans;
ll s[maxn],ma[maxn];
ll dp[maxn];

void solve()
{
    int i,j;
    ll t;
    ma[0]=dp[0]=0;
    for(i=1;i<=m;i++)
    {
        t=-INF;
        for(j=i;j<=n;j++)
        {
            dp[j]=max(dp[j-1]+s[j],ma[j-1]+s[j]);
            ma[j-1]=t;
            t=max(t,dp[j]);
        }
        ma[n]=t;
    }
}
int main()
{
    int i,j;
    while(~scanf("%I64d%I64d",&m,&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%I64d",&s[i]);
            ma[i]=0;
        }
        solve();
        printf("%I64d\n",ma[n]);
    }
    return 0;
}
/*
1 3
1 2 3
2 6
-1 4 -2 3 -2 3
2 6
-1 -2 -3 -4 -5 -6
*/