BNUOJ 14381 Wavio Sequence

Wavio Sequence

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on  UVA. Original ID: 10534
64-bit integer IO format: %lld      Java class name: Main

 

Wavio is a sequence of integers. It has some interesting properties.

  Wavio is of odd length i.e. L = 2*n + 1.

  The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.

  The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.

  No two adjacent integers are same in a Wavio sequence.

For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to find out the length of the longest Wavio sequence which is a subsequence of the given sequence. Consider, the given sequence as :

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1.

Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the output will be 9.

 

Input

The input file contains less than 75 test cases. The description of each test case is given below: Input is terminated by end of file.

 

Each set starts with a postive integer, N(1<=N<=10000). In next few lines there will be N integers.

 

Output

For each set of input print the length of longest wavio sequence in a line.

 

Sample Input

10

1 2 3 4 5 4 3 2 1 10

19

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1

5

1 2 3 4 5

Sample Output

9

9

1

解题：最长上升子序列加强版。单调队列优化！！！重点。先顺着求最长上升子序列，再逆着求最长上升子序列。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 10100;

18 int dp1[maxn],dp2[maxn],d[maxn],q[maxn];

19 int bsearch(int lt,int rt,int val) {

20     while(lt <= rt) {

21         int mid  = (lt+rt)>>1;

22         if(q[mid] < val) lt = mid+1;//严格上升单调取少于符号，上升的取少于等于

23         else rt = mid-1;

24     }

25     return lt;

26 }

27 int main() {

28     int n,i,j,head,tail;

29     while(~scanf("%d",&n)) {

30         for(i = 0; i < n; i++)

31             scanf("%d",d+i);

32         head = tail = 0;

33         for(i = 0; i < n; i++) {

34             if(head == tail) {

35                 q[head++] = d[i];

36                 dp1[i] = head-tail;

37             }else if(d[i] > q[head-1]){

38                 q[head++] = d[i];

39                 dp1[i] = head-tail;

40             }else{

41                 int it = bsearch(tail,head-1,d[i]);

42                 dp1[i] = it - tail + 1;

43                 q[it] = d[i];

44             }

45         }

46         head = tail = 0;

47         for(i = n-1; i >= 0; i--) {

48             if(head == tail) {

49                 q[head++] = d[i];

50                 dp2[i] = head-tail;

51             }else if(d[i] > q[head-1]){

52                 q[head++] = d[i];

53                 dp2[i] = head-tail;

54             }else{

55                 int it = bsearch(tail,head-1,d[i]);

56                 dp2[i] = it - tail + 1;

57                 q[it] = d[i];

58             }

59         }

60         int ans = 1;

61         for(i = 0; i < n; i++){

62             ans = max(ans,min(dp1[i],dp2[i])*2-1);

63         }

64         printf("%d\n",ans);

65     }

66     return 0;

67 }

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