HDU_1024_dp

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24252    Accepted Submission(s): 8312

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.

 

 

n个数，求m个不相交区间子段的最大值。

 

下面参考：http://www.cnblogs.com/kuangbin/archive/2011/08/04/2127085.html

状态dp[i][j]

有前j个数，组成i组的和的最大值。

决策： 第j个数，是在第包含在第i组里面，还是自己独立成组。

方程 dp[i][j]=Max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j

空间复杂度，m未知，n<=1000000，  继续滚动数组。

时间复杂度 n^3. n<=1000000.  显然会超时，继续优化。

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  下次计算的时候可以用，这样时间复杂度为 n^2.

#include<cstdio>
#include<iostream>
#include<malloc.h>
#include<cstring>
#include<map>
#include<string>
using namespace std;
#define LL long long
#define INF 0x7fffffff

int value[];
LL dp[];
LL maxn[];

int main()
{
    int m,n,num;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(dp,,sizeof(dp));
        memset(maxn,,sizeof(maxn));
        for(int i=; i<=n; i++)
            scanf("%d",&value[i]);
        LL ans=;
        LL temp;
        for(int i=; i<=m; i++)
        {
            temp=-INF;
            for(int j=i; j<=n; j++)
            {
                dp[j]=max(dp[j-],maxn[j-])+value[j];
                maxn[j-]=temp;
                temp=max(temp,dp[j]);
            }
            //cout<<dp[0][3]<<endl;

        }printf("%I64d\n",temp);
    }
    return ;
}