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;}/**101 2 3 4 5 4 3 2 1 10191 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 151 2 3 4 5**/