N – Longest Ordered SubsequenceTime Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uSubmit Status Practice POJ 2533DescriptionA 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.InputThe 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 <= 1000OutputOutput file must contain a single integer – the length of the longest ordered subsequence of the given sequence.Sample Input71 7 3 5 9 4 8Sample Output
4
#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<set>#include<map>#include<vector>#include<stack>#include<queue>#include<string>#include<cstdlib>#include<cmath>//#include<bits/stdc++.h>using namespace std;template<class T>inline T read(T&x){char c;while((c=getchar())<=32)if(c==EOF)return 0;bool ok=false;if(c=='-')ok=true,c=getchar();for(x=0; c>32; c=getchar())x=x*10+c-'0';if(ok)x=-x;return 1;}template<class T> inline T read_(T&x,T&y){return read(x)&&read(y);}template<class T> inline T read__(T&x,T&y,T&z){return read(x)&&read(y)&&read(z);}template<class T> inline void write(T x){if(x<0)putchar('-'),x=-x;if(x<10)putchar(x+'0');else write(x/10),putchar(x%10+'0');}template<class T>inline void writeln(T x){write(x);putchar('\n');}//——-ZCC IO template——const int maxn=1e4+10;const double inf=99999999;#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define For(i,t,n) for(int i=(t);i<(n);i++)typedef long long LL;typedef double DB;typedef pair<int,int> P;#define bug printf("—\n");#define mod 10007int a[maxn];int dp[maxn];void solve(){int n;scanf("%d",&n);fill(dp,dp+n,inf);for(int i=0;i<n;i++){scanf("%d",&a[i]);}for(int i=0;i<n;i++){*lower_bound(dp,dp+n,a[i])=a[i];}printf("%d\n",lower_bound(dp,dp+n,inf)-dp);}int main(){solve();return 0;}
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