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 = ;
 const    int    INF = 0x7ffffff0;
 long long     DP[SIZE][];
 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,,sizeof(DP));
         ans = -INF;
         i_0 = ;
         i_1 = ;

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