POJ 2533

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 28473 Accepted: 12361

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

71 7 3 5 9 4 8

Sample Output

4

Source

Northeastern Europe 2002, Far-Eastern Subregion

最裸的LIS了 ，直接1Y秒掉

#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <cstring>#include <map>#include <string>#include <stack>#include <cctype>#include <vector>#include <queue>#include <set>#include <iomanip>using namespace std;//#define Online_Judge#define outstars cout << "***********************" << endl;#define clr(a,b) memset(a,b,sizeof(a))#define lson l , mid  , rt << 1#define rson mid + 1 , r , rt << 1|1#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)#define mk make_pairconst int inf = 1 << 30;const int MAXN = 1000 + 150;const int maxw = 100 + 20;const int MAXNNODE = 1000 +10;const long long LLMAX = 0x7fffffffffffffffLL;const long long LLMIN = 0x8000000000000000LL;const int INF = 0x7fffffff;const int IMIN = 0x80000000;#define eps 1e-8#define mod 20071027typedef long long LL;const double PI = acos(-1.0);typedef double D;typedef pair<int , int> pii;const D e = 2.718281828459;int dp[MAXN];int main(){    //ios::sync_with_stdio(false);#ifdef Online_Judge    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);#endif // Online_Judge    int n , num;    while(~scanf("%d", & n))    {        dp[0] = -1;        int top = 0;        FOR(i , 0,  n)        {            cin >>num;            if(num > dp[top])            {                dp[++top] = num;            }            else            {                int R = top , L = 1 ;                while(L <= R)                {                    int mid = (L + R)/ 2;                    if(num > dp[mid])L = mid + 1;                    else R = mid - 1;                }                dp[L] = num;            }        }        printf("%d\n" , top);    }    return 0;}