poj 2533 Longest Ordered Subsequence(dp)

Longest Ordered Subsequence

Time Limit:2000MSMemory Limit:65536K

Total Submissions:36159Accepted:15882

Description

A numeric sequence ofaiis ordered ifa1<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

71 7 3 5 9 4 8

Sample Output

4

Source

, Far-Eastern Subregion

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#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<string>#include<algorithm>#include<cstdlib>#include<set>#include<queue>#include<stack>#include<vector>#include<map>#define N 10100#define Mod 10000007#define lson l,mid,idx<<1#define rson mid+1,r,idx<<1|1#define lc idx<<1#define rc idx<<1|1const double EPS = 1e-11;const double PI = acos ( -1.0 );const double E = 2.718281828;typedef long long ll;const int INF = 1000010;using namespace std;int dp[N];int n;int a[N];int main() {while(cin>>n) {for(int i=0; i<n; i++)scanf("%d",&a[i]);int ans=-1;for(int i=0; i<n; i++) {dp[i]=1;for(int j=0; j<=i; j++)if(a[i]>a[j])dp[i]=max(dp[j]+1,dp[i]);ans=max(ans,dp[i]);}cout<<ans<<endl;}return 0;}

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poj 2533 Longest Ordered Subsequence(dp)

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