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

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5335    Accepted Submission(s): 1939

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

Author

shǎ崽@HDU

 

Source

HDOJ Monthly Contest – 2010.06.05

 

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 #include<iostream>
 #include<stdio.h>
 #include<cstdlib>
 #include<cstring>
 #include<cstdlib>
 using namespace std;

 int a[],s[];
 int head,tail,len,n,k;
 typedef struct
 {
     int sum;
     int s,e;
 }Queue;
 Queue q[],tom,tmp;

 void Init()
 {
     int i;
     for(i=;i<=n;i++)
         scanf("%d",&a[i]);
     len=n+k;
     for(i=n+;i<=len;i++)
         a[i]=a[i-n];
     for(s[]=,i=;i<=len;i++)
         s[i]=a[i]+s[i-];
     n=n+k;
     len=len-k;
 }
 int main()
 {
     int T,i;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&k);
         Init();
         head=;tail=;
         tom.sum=s[];tom.s=;tom.e=;
         q[]=tom;
         for(i=;i<=n;i++)
         {
             tmp.sum=s[i];
             tmp.s=;
             tmp.e=i;
             while( head<=tail && q[tail].sum>tmp.sum ) tail--;
             q[++tail]=tmp;
             while( head<=tail && q[head].e+k<tmp.e ) head++;

             if(tmp.sum-q[head].sum>tom.sum && tmp.e!=q[head].e)
             {
                 tom.sum=tmp.sum-q[head].sum;
                 tom.s=q[head].e+;
                 tom.e=tmp.e;
             }
             else if( i<=k && tmp.sum>tom.sum)
             {
                 tom=tmp;
             }
         }
         printf("%d",tom.sum);
         if( tom.s>len ) tom.s-=len;
         if( tom.e>len ) tom.e-=len;
         printf(" %d %d\n",tom.s,tom.e);
     }
     return ;
 }