hdu 3415 單調佇列

_rabbit發表於2014-05-13
佇列

Max Sum of Max-K-sub-sequence

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




7 1 3
7 1 3
7 6 2
-1 1 1




 


Author


shǎ崽@HDU





求長度不超過k的最大連續子序列。


維護字首和，把字首和加入單調佇列，對於每一個下標，查詢單調佇列裡的最小值，然後做差就可以得到以這個下標結尾的最優解。


程式碼：


/* ***********************************************
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--){//維護字首和。
		 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++;//彈出距離i大於m的點。
			 if(sum[i]-sum[que[head]]>ans){//對於以i結尾的所有序列中，找單調佇列中最小的一個元素做差，這樣就可以得到以這個元素為結尾的最大和。
				 ans=sum[i]-sum[que[head]];
				 start=que[head]+1;end=i;
			 }
			 while(head<tail&&sum[que[tail-1]]>sum[i])tail--;//維護一個遞增的單調佇列。
			 que[tail++]=i;
		 }
		 if(start>n)start-=n;
		 if(end>n)end-=n;
		 printf("%I64d %I64d %I64d\n",ans,start,end);
	 }
     return 0;
}



            




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