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[200004],s[200004];
int head,tail,len,n,k;
typedef struct
{
    int sum;
    int s,e;
}Queue;
Queue q[200004],tom,tmp;

void Init()
{
    int i;
    for(i=1;i<=n;i++)
        scanf("%d",&a[i]);
    len=n+k;
    for(i=n+1;i<=len;i++)
        a[i]=a[i-n];
    for(s[0]=0,i=1;i<=len;i++)
        s[i]=a[i]+s[i-1];
    n=n+k;
    len=len-k;
}
int main()
{
    int T,i;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        Init();
        head=0;tail=0;
        tom.sum=s[1];tom.s=1;tom.e=1;
        q[0]=tom;
        for(i=2;i<=n;i++)
        {
            tmp.sum=s[i];
            tmp.s=1;
            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+1;
                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
",tom.s,tom.e);
    }
    return 0;
}