Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5690 Accepted Submission(s): 2059

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

Author

shǎ崽@HDU

求长度不超过k的最大连续子序列。

维护前缀和，把前缀和增加单调队列，对于每个下标，查找单调队列里的最小值，然后做差就能够得到以这个下标结尾的最优解。

代码：

/* ***********************************************

Author :_rabbit

Created Time :2014/5/13 2:06:57

File Name :C.cpp

************************************************ */

#pragma comment(linker, "/STACK:102400000,102400000")

#include <stdio.h>

#include <iostream>

#include <algorithm>

#include <sstream>

#include <stdlib.h>

#include <string.h>

#include <limits.h>

#include <string>

#include <time.h>

#include <math.h>

#include <queue>

#include <stack>

#include <set>

#include <map>

using namespace std;

#define INF 1LL<<60

#define eps 1e-8

#define pi acos(-1.0)

typedef __int64 ll;

ll a[201000],sum[200100],que[200100];

int main()

{

     //freopen("data.in","r",stdin);

     //freopen("data.out","w",stdout);

     ll m,n,T;

	 cin>>T;

	 while(T--){//维护前缀和。

scanf("%I64d%I64d",&n,&m);

		 for(ll i=1;i<=n;i++)scanf("%I64d",&a[i]),a[i+n]=a[i];

		 sum[0]=0;for(ll i=1;i<=2*n;i++)sum[i]=sum[i-1]+a[i];

		 ll ans=-INF,start,end;

		 ll head=0,tail=0,p;que[tail++]=0;

		 for(ll i=1;i<=2*n;i++){

			 p=max(0LL,i-m);

			 while(que[head]<p&&head<tail)head++;//弹出距离i大于m的点。

if(sum[i]-sum[que[head]]>ans){//对于以i结尾的全部序列中，找单调队列中最小的一个元素做差，这样就能够得到以这个元素为结尾的最大和。

				 ans=sum[i]-sum[que[head]];

				 start=que[head]+1;end=i;

			 }

			 while(head<tail&&sum[que[tail-1]]>sum[i])tail--;//维护一个递增的单调队列。

que[tail++]=i;

		 }

		 if(start>n)start-=n;

		 if(end>n)end-=n;

		 printf("%I64d %I64d %I64d\n",ans,start,end);

	 }

     return 0;

}

巴特西

hdu 3415 单调队列

Max Sum of Max-K-sub-sequence

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