怒刷DP之 HDU 1024

Max Sum Plus Plus

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1024

Appoint description:  System Crawler  (2015-09-05)

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 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, S ₂, 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

 

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using    namespace    std;

const    int    SIZE = 1000006;
const    int    INF = 0x7ffffff0;
long long     DP[SIZE][2];
int        S[SIZE];

long long    max(long long a,long long b);
int    main(void)
{
    int        n,m,i_0,i_1;
    long long    box,ans,temp;

    while(scanf("%d%d",&m,&n) != EOF)
    {
        memset(DP,0,sizeof(DP));
        ans = -INF;
        i_0 = 0;
        i_1 = 1;

        for(int i = 1;i <= n;i ++)
            scanf("%d",&S[i]);
        for(int j = 1;j <= m;j ++)
        {
            for(int i = 1;i <= n;i ++)
            {
                DP[i][i_1] = i - 1 < j ? -INF : DP[i - 1][i_1];
                if(i == 1)
                {
                    temp = -INF;
                    for(int k = j - 1;k < i;k ++)
                        temp = max(temp,DP[k][i_0]);
                }
                else
                    temp = temp > DP[i - 1][i_0] ? temp : DP[i - 1][i_0];
                DP[i][i_1] = (temp > DP[i][i_1] ? temp : DP[i][i_1]) + S[i];
                if(j == m)
                    ans = ans > DP[i][i_1] ? ans : DP[i][i_1];
            }
            swap(i_0,i_1);
        }
        printf("%lld\n",ans);
    }
}

long long    max(long long a,long long b)
{
    return    a > b ? a : b;
}