标签：des blog io ar os sp for strong div
Longest Ordered Subsequence
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 34440 |
|
Accepted: 15128 |
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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
using namespace std;
int n,a[1005],dp[1005],ans;
int main()
{
while(scanf("%d",&n)!=EOF)
{
ans=1;
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
dp[i]=1;
for(int i=2;i<=n;i++)
for(int j=1;j<i;j++)
if(a[j]<a[i]&&dp[i]<=dp[j])
dp[i]=dp[j]+1;
for(int i=1;i<=n;i++)
if(ans<dp[i])
ans=dp[i];
printf("%d\n",ans);
}
return 0;
}
poj 2533
标签：des blog io ar os sp for strong div
原文地址：http://www.cnblogs.com/a972290869/p/4099611.html
