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.
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).
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
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
7 1 3
7 6 2
-1 1 1
这是我们集训比赛的一道题，出题人说是什么DP，坑爹啊，比赛完后一看，单调队列，DP还做不出来
然后就果断看了一下单调队列，之是参考别人的代码后才写出来的
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）
#include <stdio.h>
#include <queue>
#include <algorithm>
#include <string.h>
using namespace std; int a[111111];
int sum[211111];
const int INF = 0x3fffffff; int main()
{
int t,n,m,i,j,k,head,end;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
j = n;
sum[0] = 0;
for(i = 1; i<=n; i++)
{
scanf("%d",&a[i]);
sum[i] = sum[i-1]+a[i];//将前i项和全部存入sum数组中
}
int ans = -INF;
for(i = n+1; i<n+k;i++)
sum[i] = sum[i-1]+a[i-n];
n = n+k-1;
deque<int> Q;//双向队列
Q.clear();
for(i = 1; i<=n; i++)
{
while(!Q.empty() && sum[i-1]<sum[Q.back()])//保持队列的单调性
Q.pop_back();
while(!Q.empty() && Q.front()<i-k)//超过k的长度则消除队列前面的元素
Q.pop_front();
Q.push_back(i-1);
if(sum[i]-sum[Q.front()]>ans)//记录，sum[n]-sum[m]所得出的是n-1到m+1之间的和
{
ans = sum[i]-sum[Q.front()];
head = Q.front()+1;
end = i;
}
}
if(end>j)
end%=j;
printf("%d %d %d\n",ans,head,end);
} return 0;
}