POJ 2533 裸的LIS
Posted stul
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ 2533 裸的LIS相关的知识,希望对你有一定的参考价值。
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.
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
#include <iostream> #define M 1005 using namespace std; int arr[M],dp[M],maxn; int main() { int n; while(cin>>n){ maxn = 0; for(int i = 0; i < n; i++) cin>>arr[i]; for(int i = 0; i < n; i++) { dp[i] = 1; for(int j = 0; j < i; j++) { if(arr[j] < arr[i]) dp[i] = max(dp[i],dp[j]+1); } maxn = max(maxn,dp[i]); } cout<<maxn<<endl; } return 0; }
以上是关于POJ 2533 裸的LIS的主要内容,如果未能解决你的问题,请参考以下文章
POJ-2533.Longest Ordered Subsequence (LIS模版题)
POJ 2533 Longest Ordered Subsequence(裸LIS)