hdu3415 Max Sum of Max-K-sub-sequence

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
   
   
   
   

这道题用的是单调队列，题目给出的是序列环，所以可以在n后面补上m-1个数，因为（j,j+1,...i)的和可以用sum[i]-sum[j-1]表示，所以可以用原题可以转换成求数字长度不大于k的最大区间和。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
#define maxn 100005
#define inf 88888888
int a[maxn],s[2*maxn];
int q[2*maxn][2];
int main()
{
	int T,n,m,i,j,sum,start,end,front,rear;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		s[0]=0;
		for(i=1;i<=n;i++){
			scanf("%d",&a[i]);
			s[i]=s[i-1]+a[i];
		}
		for(i=n+1;i<=n+m-1;i++){
			s[i]=s[i-1]+a[i-n];
		}
		sum=-inf;front=1;rear=0;
		for(i=1;i<=n+m-1;i++){
			while(front<=rear && s[i-1]<=q[rear][0])rear--;//这里每次维护i-1,因为之后的i减去q[front][1]可以直接得到[j,i]的sum值 
			rear++;
			q[rear][0]=s[i-1];
			q[rear][1]=i;
			while(front<=rear && q[front][1]+m-1<i)front++;
			if(s[i]-q[front][0]>sum){
				sum=s[i]-q[front][0];
				start=q[front][1];
				end=i;
			}
		}
	    if(end>n)end=end%n;
	    printf("%d %d %d\n",sum,start,end);
	}
	return 0;
}