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): 7413    Accepted Submission(s): 2745

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

 

单调队列即保持队列中的元素单调递增（或递减）的这样一个队列，可以从两头删除，只能从队尾插入。单调队列的具体作用在于，由于保持队列中的元素满足单调性，对于上述问题中的每个j，可以用O(1)的时间找到对应的s[i]。（保持队列中的元素单调增的话，队首元素便是所要的元素了）。

维护方法：对于每个j，我们插入s[j-1]（为什么不是s[j]? 队列里面维护的是区间开始的下标，j是区间结束的下标），插入时从队尾插入。为了保证队列的单调性，我们从队尾开始删除元素，直到队尾元素比当前需要插入的元素优（本题中是值比待插入元素小，位置比待插入元素靠前，不过后面这一个条件可以不考虑），就将当前元素插入到队尾。之所以可以将之前的队列尾部元素全部删除，是因为它们已经不可能成为最优的元素了，因为当前要插入的元素位置比它们靠前，值比它们小。我们要找的，是满足(i>=j-k+1)的i中最小的s[i]，位置越大越可能成为后面的j的最优s[i]。

在插入元素后，从队首开始，将不符合限制条件(i>=j-k+1)的元素全部删除，此时队列一定不为空。（因为刚刚插入了一个一定符合条件的元素）

//2016.8.22
#include<iostream>
#include<cstdio>
#include<queue>

using namespace std;

const int N = 100005;
const int inf = 0x3f3f3f3f;
int arr[N], sum[N*2];

int main()
{
    int T, n, k, bg, ed;
    cin>>T;
    while(T--)
    {
        scanf("%d%d", &n, &k);
        sum[0] = 0;
        for(int i = 1; i <= n; i++)
        {
              scanf("%d", &arr[i]);
            sum[i] = sum[i-1]+arr[i];
        }
        for(int i = n+1; i < n+k; i++)
              sum[i] = sum[i-1]+arr[i-n];//求一个前缀和
        int ans = -inf;
        deque<int> dq;//双端队列
        for(int i = 1; i <= n+k-1; i++)
        {
            while(!dq.empty()&&sum[i-1]<sum[dq.back()])//保持单调，使队首的sum尽量小
                  dq.pop_back();
            while(!dq.empty()&&dq.front()<i-k)
                  dq.pop_front();
            dq.push_back(i-1);
            if(sum[i]-sum[dq.front()]>ans)//sum[i]-sum[dq.front()]就是子段的和
            {
                ans = sum[i]-sum[dq.front()];
                bg = dq.front()+1;
                ed = i;
            }
        }
        if(ed>n)ed %= n;
        printf("%d %d %d
", ans, bg, ed);
    }

    return 0;
}