Longest Ordered subsequence
Time Limit: 2000MS |
|
Memory Limit: 65536K |
Total Submissions: 38980 |
|
Accepted: 17119 |
Description
A Numeric sequence of
AIis ordered if
A1<
A2< ... <
an. Let the subsequence of the given numeric sequence (
A1,
A2, ...,
an) is any sequence (
Ai1,
git, ...,
AiK), where 1 <=
I1<
I2< ... <
IK<=
N. For example, sequence (1, 7, 3, 5, 9, 4, 8) have ordered Subsequences, E. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences is of length 4, e. g., (1, 3, 5, 8).
Your program, when given the numeric sequence, must find the length of its longest ordered.
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
Source
Northeastern Europe 2002, Far-eastern subregion
N*n algorithm
#include <iostream> #include <algorithm> #include <stdio.h> #include <string.h> #include < Stdlib.h> #include <queue>using namespace Std;int n;int a[1001];int dp[1010];int main () {while (scanf ("%d" , &n)!=eof) { int maxx =-1; for (int i=0;i<n;i++) { scanf ("%d", &a[i]); } for (int i=0;i<n;i++) { dp[i] = 1; for (int j=0;j<i;j++) { if (A[i]>a[j] && dp[j]+1>dp[i]) { Dp[i] = dp[j] + 1; } } if (Maxx < dp[i]) { Maxx = Dp[i]; } } printf ("%d\n", Maxx); } return 0;}
N*logn algorithm:
#include <iostream> #include <algorithm> #include <stdio.h> #include <string.h> #include < Stdlib.h> #define INF 999999using namespace Std;int n;int dp[1010];int a[1010];int res (int len,int num) { int l = 0,r = Len; while (l!=r) { int mid = (l+r) >>1; if (dp[mid] = = num) { return mid; } else if (dp[mid]<num) { L = mid + 1; } else if (dp[mid]>num) { r = Mid; } } return l;} int main () { while (scanf ("%d", &n)!=eof) { for (int i=1;i<=n;i++) { scanf ("%d", &a[i]); } int len = 1; Dp[0] =-1; for (int i=1;i<=n;i++) { dp[i] = inf; int k = res (len,a[i]); if (k = = len) { len++; } DP[K] = A[i]; } printf ("%d\n", len-1); } return 0;}
Copyright NOTICE: This article is the original blogger article, if you have special needs, please contact Bo Master qq:793977586.
POJ 2533 longest Ordered subsequence (DP longest ascending subsequence)