PKU2533 最长上升子序列 DP

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本文介绍了一种求解最长上升子序列问题的有效算法，通过NLogN的时间复杂度实现。文章提供了一个完整的C语言实现示例，展示了如何利用二分查找来更新子序列并最终确定最长上升子序列的长度。

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15397		Accepted: 6552

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

1 7 3 5 9 4 8

Sample Output

毫无疑问 , 最长上升子序列 … 当作练习 nlogn 的算法把它水掉了 …

代码如下 :

#include<stdio.h> #include<string.h> #define MAXN 1005 int i,n,a[MAXN],c[MAXN],len[MAXN],pos,max,total; int find(int a,int l,int r) { int mid=(l+r)/2; while (l<=r) { if (c[mid]==a) return mid; else { if (a>c[mid]) l=mid+1; else r=mid-1; mid=(l+r)/2; } } return l; } int main() { while (scanf("%d",&n)!=EOF) { for (i=1;i<=n;++i) scanf("%d",&a[i]); for (i=1;i<=n;++i) c[i]=20000; c[1]=a[1]; len[1]=1; total=1; max=1; for (i=1;i<=n;++i) { pos=find(a[i],1,total+1); c[pos]=a[i]; len[i]=pos; if (pos>max) {max=pos;total++;} } printf("%d/n",max); } return 0; }