hdu 1024 Max Sum Plus Plus

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25639    Accepted Submission(s): 8884

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）

 

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#include<iostream>
#include<stdio.h>
#include<string>
#include<cstring>
using namespace std;
const int maxn = ;
int dp[maxn];
int pri[maxn];
int a[maxn];

int main()
{
    int n,m;
    while(~scanf("%d%d",&m,&n))
    {
        for(int i=;i<=n;i++) scanf("%d",a+i);
        int tmp_max=-0x3fff;
        dp[]=;
        //pri[0]=0;
        memset(pri,,sizeof(pri));
        for(int i=;i<=m;i++)
        {
            tmp_max=-0x3fffffff;
            for(int j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-],pri[j-])+a[j];
                pri[j-]=tmp_max;
                if(tmp_max<dp[j]) tmp_max=dp[j];
            }

        }
        printf("%d\n",tmp_max);
    }
    return ;
}

设输入的数组为a[1...n]，从中找出m个段，使者几个段的和为最大

dp[i][j]表示前j个数中取i个段的和的最大值，其中最后一个段包含a[j]。(这很关键)

则状态转移方程为：

dp[i][j]=max{dp[i][j-1]+a[j],max{dp[i-1][t]}+a[j]}    i-1=<t<j-1

因为dp[i][j]中a[j]可能就自身一个数组成最后一段，或者a[j]与a[j-1]等前面的数组成最后一段。

此题n数据太大，二维数组开不下，而且三重循环，想到状态转移方程后还是困难重重。

想想，二维数组不行的话，肯定要压缩成一维数组：

因为dp[i-1][t]的值只在计算dp[i][j]的时候用到，那么没有必要保存所有的dp[i][j] for i=1 to m,这样我们可以用一维数组存储。

用pre[j]表示j之前一个状态dp[i-1][]中1-j之间，不一定包含a[j]的最大字段和，然后推dp[i][j]状态时，dp[i][j]=max{pre[j-1]，dp[j-1]}+a[j];

褐色的为了方便理解，其实不存在。