POJ 2533 Longest Ordered Subsequence 最长上升子序列

Longest Ordered Subsequence

Description

A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= 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

7
1 7 3 5 9 4 8

Sample Output

4

代码 ：

/*都是代码 看着烦 来点 新东西
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*/
#include<cstdio>
#include<iostream>
using namespace std ;
#define M 1010
int a[M] , b[M] ;
int find( int s , int e , int w ){ // 二分查找 比 a[] 大或者相等的然后返回下标
if( s == e ) return s ;
int mid = ( s + e ) >> 1 ;
if( b[mid] < w )  return find( mid + 1 , e , w ) ;
else return find( s , mid , w ) ;
}
int main()
{
int i  , j , n , m ;
while(scanf( "%d" , &n ) != EOF ){
memset( b , 0  , sizeof(b) ) ;
for( i = 1 ;i <= n ;i++)
scanf( "%d" , &a[i] ) ;
b[0] = -200000 ; int len = 0 ;
for( i = 1 ;i <= n ;i++){
if( a[i] > b[len] ) j = ++len ;
else j = find( 1 , len , a[i] ) ;
b[j] = a[i] ; // 替换

}
cout << len << endl ;
}
}