Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4080 Accepted Submission(s): 1453

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

 1 /*
 2 题目大意：给出一个有N个数字（N<=10^5）的环状序列，让你求一个和最大的连续子序列。这个连续子序列的长度小于等于K。
 3 输出和，起始和结束位置
 4 */
 5 #include<iostream>
 6 #include<queue>
 7 using namespace std;
 8
 9 const int INF = 0x3fffffff;
10 const int maxn = 100010;
11 int num[maxn],sum[maxn];
12
13 int main()
14 {
15     int T,i,j;
16     int N,K,n;
17     cin>>T;
18     while(T--)
19     {
20         cin>>N>>K;
21         sum[0]=0;
22         for(i=1;i<=N;i++)
23         {
24             cin>>num[i];
25             sum[i]=sum[i-1]+num[i];
26         }
27         for(i=N+1;i<N+K;i++)
28         {
29             sum[i]=sum[i-1]+num[i-N];
30         }
31         n=N+K-1;
32         deque <int> q;
33         q.clear();
34         int ans=-INF;
35         int start,end;
36         //枚举以j结尾的区间
37         for(j=1;j<=n;j++)
38         {
39             while(!q.empty() && sum[j-1]<sum[q.back()])
40                 q.pop_back();
41             while(!q.empty() && q.front()<(j-K))//区间最大长度是K ,没有等号，因为入队的是 (j-1)
42                 q.pop_front();
43             q.push_back(j-1);
44             if(sum[j]-sum[q.front()]>ans)
45             {
46                 ans=sum[j]-sum[q.front()];
47        //i到j和是,sum[8] - sum[5] = sum[6]+sum[7]+sum[8]

　　　　　　         start=q.front()+1;
48                 end=j;
49             }
50         }
51         cout<<ans<<" "<<start<<" "<<(end>N?end%N:end)<<endl;
52     }
53     return 0;
54 }////暴力枚举起点和终点估计会超时