UVA 10534 Wavio Sequence

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                                   Output for 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

9 9 1

双向求最长递增子序列，然后相加求最大值，这里要用nlogn的算法。

#include<iostream>  
#include<algorithm>
#include<math.h>
#include<cstdio>
#include<string>
#include<string.h>
using namespace std;
const int maxn = 10010;
int n, a[maxn], b[maxn], f[maxn][2], m, i, j, w;

int main(){
  while (~scanf("%d", &n))
  {
    for (i = 0; i < n; i++) scanf("%d", &a[i]);
    memset(b, 0, sizeof(b));
    for (m = 0, i = 0; i < n; i++)
    {
      int q = 0, h = m;
      while (q <= h)
      {
        int mid = (q + h) / 2;
        if (a[i]>b[mid]) q = mid;
        else h = mid - 1;
        if (a[i]>b[h]) break;
        if (q == h) break;
        if (q + 1 == h) h = q;
      }
      b[h + 1] = a[i];
      if (h == m) m++;
      f[i][0] = h + 1;
    }
    memset(b, 0, sizeof(b));
    for (m = 0, i = n - 1; i >= 0; i--)
    {
      int q = 0, h = m;
      while (q<=h)
      {
        int mid = (q + h) / 2;
        if (a[i]>b[mid]) q = mid;
        else h = mid - 1;
        if (a[i] > b[h]) break;
        if (q == h) break;
        if (q + 1 == h) h = q;
      }
      b[h + 1] = a[i];
      if (h == m) m++;
      f[i][1] = h + 1;
    }
    for (w = 0, i = 0; i < n; i++)w = max(w, 2 * min(f[i][0], f[i][1]) - 1);
    cout << w << endl;
  }
  return 0;
}