Longest Ordered Subsequence
Time Limit: 2000MS Memory Limit: 65536K
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
The main idea of the topic: Enter an N to indicate that there is a sequence of length N, and then find the length of the longest sequence in ascending order.
For example, the longest sequence of 1 7 3 5 9 4 8 is 1 3 5 9 with 4 units of length.
It can be seen that if dp[] is used to record the longest sequence to reach there, dp[1]=1. Starting from the second number, if the second number is greater than the first number num[2]>num[1] then dp[2]=dp[1]+1 otherwise dp[2]=dp[1]
#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int maxn = 1e3+5;
int num[maxn];
int dp[maxn];
int n;
int temp;
int main(){
while(~scanf("%d",&n)){
for( int i=1 ; i<=n ; i++ ){
scanf("%d",&num[i]);
}
dp[1]=1;
for( int i=2 ; i<=n ; i++ ){
temp = 0;
for( int j=1 ; j<i ; j++ ){
if(num[i]>num[j]){
if(temp<dp[j]){
temp = dp[j];
}
}
}
dp[i] = temp + 1;
}
sort(dp+1,dp+n+1);
printf("%d\n",dp[n]);
}
return 0;
}