POJ 2533 Longest Ordered Subsequence

Posted 2020-09-21 zsychanpin

链接：http://poj.org/problem?

id=2533

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 38875 Accepted: 17066

Description

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

大意——给一队排列整齐的数列ai，找到数列ai中最长上升子序列的长度。

思路——这是一道很典型的LIS问题。设dp[i]表示以第i个数为后缀的单调序列的最大长度，则dp[0]=1,dp[i+1]= max{1,dp[j]+1|j=0,1,...,i&&a[i+1]>a[j]},max{dp[i]|i=0,...,n-1}即为结果。

复杂度分析——时间复杂度：O(n^2),空间复杂度：O(n)

附上AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
using namespace std;
typedef unsigned int UI;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
const double pi = acos(-1.0);
const double e = exp(1.0);
const int maxn = 1005;
int num[maxn];
int dp[maxn];

int main()
{
	ios::sync_with_stdio(false);
	int n;
	while (scanf("%d", &n) != EOF)
	{
		for(int i=0; i<n; i++)
			scanf("%d", &num[i]);
		dp[0] = 1;
		for (int i=1; i<n; i++)
		{
			dp[i] = 1;
			for (int j=0; j<i; j++)
				if (num[j] < num[i])
					dp[i] = max(dp[i], dp[j]+1);
		}
		int ans = 0;
		for (int i=0; i<n; i++)
			ans = max(ans, dp[i]);
		printf("%d\n", ans);
	}
	return 0;
}