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。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 int dp1[maxn],dp2[maxn],d[maxn],q[maxn];
 int bsearch(int lt,int rt,int val) {
     while(lt <= rt) {
         int mid  = (lt+rt)>>;
         if(q[mid] < val) lt = mid+;//严格上升单调取少于符号，上升的取少于等于
         else rt = mid-;
     }
     return lt;
 }
 int main() {
     int n,i,j,head,tail;
     while(~scanf("%d",&n)) {
         for(i = ; i < n; i++)
             scanf("%d",d+i);
         head = tail = ;
         for(i = ; i < n; i++) {
             if(head == tail) {
                 q[head++] = d[i];
                 dp1[i] = head-tail;
             }else if(d[i] > q[head-]){
                 q[head++] = d[i];
                 dp1[i] = head-tail;
             }else{
                 int it = bsearch(tail,head-,d[i]);
                 dp1[i] = it - tail + ;
                 q[it] = d[i];
             }
         }
         head = tail = ;
         for(i = n-; i >= ; i--) {
             if(head == tail) {
                 q[head++] = d[i];
                 dp2[i] = head-tail;
             }else if(d[i] > q[head-]){
                 q[head++] = d[i];
                 dp2[i] = head-tail;
             }else{
                 int it = bsearch(tail,head-,d[i]);
                 dp2[i] = it - tail + ;
                 q[it] = d[i];
             }
         }
         int ans = ;
         for(i = ; i < n; i++){
             ans = max(ans,min(dp1[i],dp2[i])*-);
         }
         printf("%d\n",ans);
     }
     return ;
 }