HDu 3415 Max Sum of Max-K-sub-sequence単調キュー.

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子@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
",tom.s,tom.e);

65     }

66     return 0;

67 }