POJ 2533 Longest Ordered Subsequence

Longest Ordered Subsequence

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2533
64-bit integer IO format: %lld      Java class name: Main

 

A 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.

 

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

 

解题：LIS

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1010;
int dp[maxn],d[maxn],n;
int main() {
    while(~scanf("%d",&n)){
        for(int i = 0; i < n; ++i){
            scanf("%d",d+i);
            dp[i] = INF;
        }
        dp[n] = INF;
        for(int i = 0; i < n; ++i){
            int *p = lower_bound(dp,dp+n+1,d[i]);
            *p = d[i];
        }
        printf("%d
",lower_bound(dp,dp+n+1,INF)-dp);
    }
    return 0;
}