HDU 1024 Max Sum Plus Plus [动态规划+m子段和的最大值]

Max Sum Plus Plus

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

Total Submission(s): 22262    Accepted Submission(s): 7484

 

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

[题意]:输入一个m，n分别表示成m组，一共有n个数即将n个数分成m组，m组的和加起来得到最大值并输出。

[分析]:

状态dp[i][j]表示前j个数分成i组的最大值。

动态转移方程：dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) (0<k<j)

dp[i][j-1]+a[j]表示的是前j-1分成i组，第j个必须放在前一组里面。

max( dp[i-1][k] ) + a[j] )表示的前（0<k<j）分成i-1组，第j个单独分成一组。

但是题目的数据量比较到，时间复杂度为n^3，n<=1000000，显然会超时，继续优化。

max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个
的最大值 用数组保存下来 ，这样时间复杂度为 n^2。

 

[代码]:

/*
输入一个m，n分别表示成m组,一共有n个数
即将n个数分成m组，
m组的和加起来得到最大值并输出。
*/
#include <bits/stdc++.h>
using namespace std;
const int N=1000000;
#define INF 0x7fffffff

int a[N+10];
int dp[N+10],Max[N+10];

int main()
{
    int n,m,maxs;
    while(~scanf("%d%d",&m,&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        memset(Max,0,sizeof(Max));

        for(int i=1;i<=m;i++)
        {
            maxs=-INF;
            for(int j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1]+a[j], Max[j-1]+a[j]);
                Max[j-1]=maxs;
                maxs=max(maxs, dp[j]);
            }
        }
        printf("%d
",maxs);
    }
}