Uva 10534 波浪串 --> LIS变形(n*logn)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1475

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

题目大意：找出一个序列长度n*2+1,前n个是递增序列，后n个事递减序列（不能有相等的）求最大的长度。

解题思路：

             对于每个位置找出其最长上升子序列和反向的最长上升子序列，然后取二者的小者。而后枚举i，找最长的。注意要用前面提到的n*logn的复杂度算法来求LIS否则会超时的。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
int g[10005],a[10005],d[10005],n,sum1[10005],sum2[10005];
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1; i<=n; i++)
            g[i]=INF;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
            int k=lower_bound(g+1,g+n+1,a[i])-g;
            sum1[i]=k;
            g[k]=a[i];
        }
        for(int i=1; i<=n; i++)
            g[i]=INF;
        for(int i=n-1; i>=0; i--)
        {
            int k=lower_bound(g+1,g+n+1,a[i])-g;
            sum2[i]=k;
            g[k]=a[i];
        }
        /*
        for(int i=0;i<n;i++)
            printf("%d %d\n",sum1[i],sum2[i]);
            */
        int maxx=-1;
        for(int i=0; i<n; i++)
            maxx=max(maxx,min(sum1[i],sum2[i]));
        printf("%d\n",maxx*2-1);
    }
    return 0;
}
/**

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

**/