F - Longest Ordered Subsequence POJ - 2533

Longest Ordered Subsequence Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 49504 Accepted: 21978

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

7 1 7 3 5 9 4 8

Sample Output

4

Source

Northeastern Europe 2002, Far-Eastern Subregion 最简单的最长递增子序列，wa了很多次，后来明白结果不一定是dp{ n-1 ]，应该所有的最大值，忽略了。。谨记！！

#include<stdio.h> #include<string.h> #include<math.h> #include<string> #include<algorithm> #include<queue> #include<vector> #include<map> #include<set> typedef long long ll; using namespace std; int main() { int ans,dp[10005],n,a[10005],i,j; scanf("%d",&n); for(i=0;i<n;i++) scanf("%d",&a[i]),dp[i]=1; for(i=1;i<n;i++) { for(j=i-1;j>=0;j--) { if(a[i]>a[j]) { dp[i]=max(dp[i],dp[j]+1); } } } ans=dp[0]; for(i=1;i<n;i++) ans=max(ans,dp[i]); printf("%d\n",ans); return 0; }