hdu 3415 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): 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; }

转载于:https://www.cnblogs.com/jzssuanfa/p/6798497.html