Max Sum of Max-K-sub-sequence 单调队列（有技巧）

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（有技巧）

***************************************************************************************************************************

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define  inf  0x7fffffff
using namespace std;
int a[100011],sum[201001];
int que[201001];
int cas,n,k;
int st,en,result;
void solve()
{
    int head=1,tail=0;
    result=-inf;
    st=inf;
    int it;
    for(it=1;it<=n+k;it++)
    {
        while(head<=tail&&sum[it-1]<sum[que[tail]])//不满足条件，从尾部弹出
         tail--;
        while(head<=tail&&que[head]<it-k)//如果超出范围，从顶部弹出
         head++;
        tail++;
        que[tail]=it-1;//此处压值时有技巧
        if(sum[it]-sum[que[head]]>result)//每次更新极优值
        {
            result=sum[it]-sum[que[head]];
            st=que[head]+1;
            en=it;
        }
    }
    if(en>n)
     en-=n;
}
int main()
{
    int i,j;
    scanf("%d",&cas);
    while(cas--)
    {
        sum[0]=0;
        scanf("%d %d",&n,&k);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        for(i=n+1;i<=n+k;i++)
         sum[i]=sum[i-1]+a[i-n];
        solve();
        printf("%d %d %d
",result,st,en);

    }
}