Longest Ordered Subsequence（POJ-2533）

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Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 49939 Accepted: 22173

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_{i_K}), 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

题目大意：

给一个序列，求序列中的最长递增子序列。

解题思路：

DP的经典题型，dp数组记录当前最长子序列的长度，边输入边判断，伪状态转移方程：DP[ i ] = DP(之前的比 i 小的而且dp值最大的)+1;

代码如下：

#include<stdio.h>int dp[1003];int a[1003];int main(){int n,m;while(~scanf("%d",&n)){int answer = 0;for(int i=1 ; i<=n ; i++){scanf("%d",&a[i]);int t_m = 0;for(int j=1 ; j<i ; j++){if(a[j]<a[i] && dp[j]>t_m)t_m = dp[j];}dp[i] = t_m+1;if(dp[i]>answer) answer = dp[i];}printf("%d\n",answer);}return 0;}

题目传送门。。。。。

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