Max Sum of Max-K-sub-sequence 单调队列（有技巧）

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

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单调队列（有技巧）

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 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<algorithm>
 7 #define  inf  0x7fffffff
 8 using namespace std;
 9 int a[100011],sum[201001];
10 int que[201001];
11 int cas,n,k;
12 int st,en,result;
13 void solve()
14 {
15     int head=1,tail=0;
16     result=-inf;
17     st=inf;
18     int it;
19     for(it=1;it<=n+k;it++)
20     {
21         while(head<=tail&&sum[it-1]<sum[que[tail]])//不满足条件，从尾部弹出
22          tail--;
23         while(head<=tail&&que[head]<it-k)//如果超出范围，从顶部弹出
24          head++;
25         tail++;
26         que[tail]=it-1;//此处压值时有技巧
27         if(sum[it]-sum[que[head]]>result)//每次更新极优值
28         {
29             result=sum[it]-sum[que[head]];
30             st=que[head]+1;
31             en=it;
32         }
33     }
34     if(en>n)
35      en-=n;
36 }
37 int main()
38 {
39     int i,j;
40     scanf("%d",&cas);
41     while(cas--)
42     {
43         sum[0]=0;
44         scanf("%d %d",&n,&k);
45         for(i=1;i<=n;i++)
46         {
47             scanf("%d",&a[i]);
48             sum[i]=sum[i-1]+a[i];
49         }
50         for(i=n+1;i<=n+k;i++)
51          sum[i]=sum[i-1]+a[i-n];
52         solve();
53         printf("%d %d %d\n",result,st,en);
54 
55     }
56 }

View Code

转载于:https://www.cnblogs.com/sdau--codeants/p/3450678.html