hdu3415(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): 10481    Accepted Submission(s): 3853

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

用sum数组存前缀和，a【i.....j]=sum[j]-sum[i-1];max(a[1....j])=sum[j]-min(sum[i-1])    j-(i-1)<=k;

用单调栈维护最小的sum[i-1]即可；

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m;int a[100005],sum[200100],q[200100];
int start,endi,total;int res;int ans;
void get()
{
	int head=0,tail=0;
	 ans=-100000000;
	for(int i=1;i<total;i++)
	{
		while(head<tail&&sum[q[tail-1]]>=sum[i-1])tail--;
		while(head<tail&&q[head]<i-m)head++;
		q[tail++]=i-1;
		if(sum[i]-sum[q[head]]>ans)
		{
			start=q[head];
			endi=i;
			ans=sum[endi]-sum[start];
		}
	}
} 
int main()
{
	int t;scanf("%d",&t);
	while(t--)
	{
		cin>>n>>m;sum[0]=0;
		for(int i=1;i<=n;i++)
		{
			cin>>a[i];
			sum[i]=sum[i-1]+a[i];
		}
		for(int i=1;i<m;i++)
		sum[n+i]=sum[n+i-1]+a[i];
		total=n+m;
		get();
		start++;
		if(start>n)start-=n;
		if(endi>n)endi-=n;
		 cout<<ans<<" "<<start<<" "<<endi<<endl;
	}
	return 0;
}