hdu 3415 Max Sum of Max-K-sub-sequence 单调队列dp

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4581    Accepted Submission(s): 1656

Problem Description

Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1].
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

 

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. 
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the minimum length of them.

 

Sample Input

   
   
   
   

    
    
    
    4
6 3
6 -1 2 -6 5 -5
6 4
6 -1 2 -6 5 -5
6 3
-1 2 -6 5 -5 6
6 6
-1 -1 -1 -1 -1 -1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    7 1 3
7 1 3
7 6 2
-1 1 1
   
   
   
   

 

-----------------------

单调队列优化dp的入门题

-----------------------

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int a[411111];
int f[411111];
int que[1111111];
int pt[1111111];
int n,k;
int T;
int head,tail;
int sum[411111];
int max_sum,start,end;

int main()
{
    scanf("%d",&T);
    while (T--)
    {
        memset(f,0,sizeof(f));
        memset(que,0,sizeof(que));
        memset(pt,0,sizeof(pt));
        memset(sum,0,sizeof(sum));
        scanf("%d%d",&n,&k);
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i+n]=a[i];
        }
        for (int i=1;i<=n+k;i++)
        {
            sum[i]+=sum[i-1]+a[i];
        }
        //f[i]=max(sum[i]-sum[k]);
        head=tail=0;
        max_sum=start=end=-1e9;
        for (int i=1;i<=n+k;i++)
        {
            while ((head<tail)&&(i-pt[head]>k)) head++;
            while ((head<tail)&&(sum[i-1]<=que[tail-1])) tail--;
            que[tail]=sum[i-1],pt[tail++]=i-1;
            f[i]=sum[i]-que[head];
            if (f[i]>max_sum)
            {
                max_sum=f[i];
                start=pt[head]+1;
                end=i;
            }
        }
        if (start>n) start=start-n;
        if (end>n) end=end-n;
        printf("%d %d %d\n",max_sum,start,end);
    }
    return 0;
}