poj 2823 Sliding Window 单调队列

Sliding Window Time Limit: 12000MS   Memory Limit: 65536K Total Submissions: 35031   Accepted: 10343 Case Time Limit: 5000MS Description An array of size   n  ≤ 10 6  is given to you. There is a sliding window of size   k  which is moving from the very left of the array to the very right. You can only see the   k  numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:   The array is   [1 3 -1 -3 5 3 6 7], and   k  is 3. Window position Minimum value Maximum value [1  3  -1] -3  5  3  6  7  -1 3  1 [3  -1  -3] 5  3  6  7  -3 3  1  3 [-1  -3  5] 3  6  7  -3 5  1  3  -1 [-3  5  3] 6  7  -3 5  1  3  -1  -3 [5  3  6] 7  3 6  1  3  -1  -3  5 [3  6  7] 3 7 Your task is to determine the maximum and minimum values in the sliding window at each position.  Input The input consists of two lines. The first line contains two integers   n  and   k  which are the lengths of the array and the sliding window. There are   n  integers in the second line.   Output There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.   Sample Input 8 3 1 3 -1 -3 5 3 6 7 Sample Output -1 -3 -3 -3 3 3 3 3 5 5 6 7 Source POJ Monthly--2006.04.28, Ikki

[Submit]   [Go Back]   [Status]   [Discuss]

Home Page   Go Back  To top

 

 

 1 /*

 2    单调队列基础题。

 3 */

 4 /*

 5 单调队列维护。

 6 以单调递减队列为例。10 9 8 7 6 ...

 7 初始化

 8 head = 0; tail = -1;

 9 进队列：从后往前查找。满足  while( head<=tail  && q[ tail ].rp < tmp.rp ) tail --;

10                                             此时相当于舍弃 tail 原来保存到的数据。

11 出队列：从前往后查找。满足  while( head<=tail  && q[ tail ].num <= conunt ) head ++;

12                                             此时相当于避免比当前值先进来，却又比较大的值，被输出。

13 */

14 

15 #include<iostream>

16 #include<stdio.h>

17 #include<cstring>

18 #include<cstdlib>

19 using namespace std;

20 

21 int a[1000002];

22 typedef struct

23 {

24     int num;

25     int rp;

26 }Queue;

27 Queue q[1000002];

28 int main()

29 {

30     int n,i,k,flag;

31     int Num,conunt,head,tail;

32     Queue tmp;

33     while(scanf("%d%d",&n,&k)>0)

34     {

35         Num=0;

36         conunt=0;

37         head = 0; tail = -1;

38         flag=1;

39 

40         for(i=1;i<=n;i++)

41         {

42             scanf("%d",&a[i]);

43             tmp.rp=a[i];

44             tmp.num=++Num;

45             while( head<=tail && q[tail].rp > tmp.rp)

46                 tail--;

47             q[++tail]=tmp;

48             if(i>=k)

49             {

50                 while( head<=tail && q[head].num<=conunt)

51                     head ++;

52                 if(head>tail) ;

53                 else if(flag==1)

54                 {

55                     printf("%d",q[head].rp);

56                     flag=0;

57                 }

58                 else printf(" %d",q[head].rp);

59                 conunt++;

60             }

61         }

62         printf("\n");

63         Num=0;

64         conunt=0;

65         head = 0; tail = -1;

66         flag=1;

67         for(i=1;i<=n;i++)

68         {

69             tmp.rp=a[i];

70             tmp.num=++Num;

71             while( head<=tail && q[tail].rp < tmp.rp)

72                 tail--;

73             q[++tail]=tmp;

74             if(i>=k)

75             {

76                 while( head<=tail && q[head].num<=conunt)

77                     head ++;

78                 if(head>tail) ;

79                 else if(flag==1)

80                 {

81                     printf("%d",q[head].rp);

82                     flag=0;

83                 }

84                 else printf(" %d",q[head].rp);

85                 conunt++;

86             }

87         }

88         printf("\n");

89     }

90     return 0;

91 }