Max Sum Plus Plus HDU - 1024

最新推荐文章于 2024-07-21 14:30:00 发布

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本文介绍了一个基于动态规划的算法，用于解决给定序列中寻找特定数量的最大子段和的问题。通过逐个考虑序列中的元素，并维护当前最优解的状态，该算法能够在较大的输入规模下高效地找到答案。

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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, S₂, 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

Hint

Huge input, scanf and dynamic programming is recommended.

又到了被动态规划虐脑的时候了。。

崩溃T^T

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int INF=0x3f3f3f3f;
int a[1123456];
int dp[1123456];
int mm[1123456];
int main()
{
    int n,m;
    while(~scanf("%d%d",&m,&n))
    {
        int i,j;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            mm[i]=0;
            dp[i]=0;
        }
        dp[0]=0;
        mm[0]=0;
        int ma;
        for(i=1;i<=m;i++)
        {
            ma=-INF;
            for(j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1]+a[j],mm[j-1]+a[j]);//dp[j-1]表示的是以j-1结尾i个子段的和..mm[j-1]表示前j-1个元素i-1个子段的和
                mm[j-1]=ma;//保证a[i]是一个独立的子段
                ma=max(ma,dp[j]);
            }
        }
        printf("%d\n",ma);
    }
    return 0;
}