hdu 3415 Max Sum of Max-K-sub-sequence 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): 5335    Accepted Submission(s): 1939

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

 

 

Source

HDOJ Monthly Contest – 2010.06.05

 

 

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 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cstdlib>
 6 using namespace std;
 7 
 8 int a[200004],s[200004];
 9 int head,tail,len,n,k;
10 typedef struct
11 {
12     int sum;
13     int s,e;
14 }Queue;
15 Queue q[200004],tom,tmp;
16 
17 void Init()
18 {
19     int i;
20     for(i=1;i<=n;i++)
21         scanf("%d",&a[i]);
22     len=n+k;
23     for(i=n+1;i<=len;i++)
24         a[i]=a[i-n];
25     for(s[0]=0,i=1;i<=len;i++)
26         s[i]=a[i]+s[i-1];
27     n=n+k;
28     len=len-k;
29 }
30 int main()
31 {
32     int T,i;
33     scanf("%d",&T);
34     while(T--)
35     {
36         scanf("%d%d",&n,&k);
37         Init();
38         head=0;tail=0;
39         tom.sum=s[1];tom.s=1;tom.e=1;
40         q[0]=tom;
41         for(i=2;i<=n;i++)
42         {
43             tmp.sum=s[i];
44             tmp.s=1;
45             tmp.e=i;
46             while( head<=tail && q[tail].sum>tmp.sum ) tail--;
47             q[++tail]=tmp;
48             while( head<=tail && q[head].e+k<tmp.e ) head++;
49 
50             if(tmp.sum-q[head].sum>tom.sum && tmp.e!=q[head].e)
51             {
52                 tom.sum=tmp.sum-q[head].sum;
53                 tom.s=q[head].e+1;
54                 tom.e=tmp.e;
55             }
56             else if( i<=k && tmp.sum>tom.sum)
57             {
58                 tom=tmp;
59             }
60         }
61         printf("%d",tom.sum);
62         if( tom.s>len ) tom.s-=len;
63         if( tom.e>len ) tom.e-=len;
64         printf(" %d %d\n",tom.s,tom.e);
65     }
66     return 0;
67 }

 

转载于:https://www.cnblogs.com/tom987690183/p/3558256.html