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): 16803    Accepted Submission(s): 5525

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 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 _xor 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


   
   

    
    

     
     Hint
    
    
Huge input, scanf and dynamic programming is recommended.

   
   
   
    
  
  

 

Author

JGShining（极光炫影）

解题思路：求n个数的m个不相交子区间的和的最大值，令dp[i][j]表示以第i个数为结尾的j个区间的最大值，

dp[i][j]=max(dp[i-1][j]+arr[i],max(dp[k][j-1])+arr[i],0<k<i)；

注意到dp[i][j]只与dp[i-1][j]和dp[k][j-1]有关，采用滚动数组优化

同时dp[k][j-1]可以在之前的0---i-1中计算出来用数组保存下

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <list>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <bitset>
#include <cmath>
#include <utility>
#define Maxn 1000005
#define Maxm 1000005
#define lowbit(x) x&(-x)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define PI acos(-1.0)
#define make_pair MP
#define LL long long 
#define Inf (1LL<<62)
#define inf 0x3f3f3f3f
#define re freopen("in.txt","r",stdin)
#define wr freopen("out.txt","w",stdout)
using namespace std;
int arr[Maxn];
int dp[Maxn];
int dpp[Maxn];
int read()
{
    int res=0;
    bool flag=true;
    while(1)
    {
         char ch=getchar();
         if(ch>='0'&&ch<='9')
             res=res*10+ch-'0';
         else if(ch=='-')
         {
            flag=false;
            continue;
         }
         else
             break;
    }
    return flag?res:-res;
}
int main()
{
    int n,m;
    int ans;
    //re;wr;
    while(~scanf("%d%d",&m,&n))
    {
        getchar();
        for(int i=0;i<=n;i++)
            arr[i]=dp[i]=dpp[i]=0;
        for(int i=1;i<=n;i++)
            arr[i]=read();
        for(int j=1;j<=m;j++)
        {
            ans=-inf;
            for(int i=j;i<=n;i++)
                //下标要从j开始，这样才能保证滚动数组是正确的
            {
                dp[i]=max(dpp[i-1],dp[i-1])+arr[i];//dp[i][j]=max(dp[i-1][j]+arr[i],max(dp[k][j-1])+arr[i],0<k<i)
                dpp[i-1]=ans;//dpp存储max(dp[k][j-1]),0<k<i
                ans=max(ans,dp[i]);//ans表示以i为结尾的选择了j段的最大值
            }
        }
        printf("%d\n",ans);

    }
    return 0;
}