Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 38980		Accepted: 17119

Description

A numeric sequence of 

a_i is ordered if 

a₁ < 

a₂ < ... < 

a_N. Let the subsequence of the given numeric sequence (

a₁, 

a₂, ..., 

a_N) be any sequence (

a_i1, 

a_i2, ..., 

a_iK), where 1 <= 

i₁ < 

i₂ < ... < 

i_K <= 

N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

71 7 3 5 9 4 8

Sample Output

4

Source

, Far-Eastern Subregion

n*n算法

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            using namespace std;int n;int a[1001];int dp[1010];int main(){ while(scanf("%d",&n)!=EOF) { int maxx = -1; for(int i=0;i
            
             a[j] && dp[j]+1>dp[i]) { dp[i] = dp[j] + 1; } } if(maxx < dp[i]) { maxx = dp[i]; } } printf("%d\n",maxx); } return 0;}
            
           
          
         
        
       
      

n*logn算法：

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #define inf 999999using namespace std;int n;int dp[1010];int a[1010];int res(int len,int num){    int l = 0,r = len;    while(l!=r)    {        int mid = (l+r)>>1;        if(dp[mid] == num)        {            return mid;        }        else if(dp[mid]
           
            num) { r = mid; } } return l;}int main(){ while(scanf("%d",&n)!=EOF) { for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } int len = 1; dp[0] = -1; for(int i=1;i<=n;i++) { dp[i] = inf; int k = res(len,a[i]); if(k == len) { len++; } dp[k] = a[i]; } printf("%d\n",len-1); } return 0;}