N - Longest Ordered Subsequence——POJ 最长递增子序列

N - Longest Ordered Subsequence

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

最长递增子序列：简单一维DP，DP[i]表示前n个数的最长子序列！

#include<stdio.h>#include<string.h>#include<math.h>int my_max(int x,int y)  {return x>y?x:y;}int a[2005],dp[2005];int main(){int n,i,j,k;while(scanf("%d",&n)!=EOF){memset(dp,0,sizeof(dp));memset(a,0,sizeof(a));for(i=1;i<=n;i++)scanf("%d",&a[i]);for(i=1;i<=n;i++)dp[i]=1;int max=1;for(i=2;i<=n;i++){for(j=1;j<i;j++){if(a[i]>a[j]){dp[i]=my_max(dp[i],dp[j]+1);}}max=my_max(dp[i],max);}printf("%d\n",max);}return 0;}