poj 2533 Longest Ordered Subsequence

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35524		Accepted: 15582

DescriptionA 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.InputThe 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 <= 1000OutputOutput 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

Memory: 704K

Time: 47MS

Language: G++

User: Morezy

#include <iostream>using namespace std;int array[1050];int liss[1050];//保存以当前数为最大的数(在末尾)的最长递增子序列的长度int pre[1050];//当前数的前驱的位置int LIS(int *a, int len){int i,j,max;for (i = 0; i < len; i++)//初始化，最长递增子序列长度最小是1，即每个序列都是它们本身{pre[i] = i;liss[i] = 1;}for (i = 1,max = 1; i < len; i++)//以数组第二个开始{for(j = 0; j < i; j++){if (array[j] < array[i] && liss[j] + 1 > liss[i]){liss[i] = liss[j]+1;pre[i] = j;}if (liss[i] > max){max = liss[i];}}}return max;}int main(){int n,i;while (cin >> n && n){for (i = 0; i < n; i++){cin >> array[i];}cout << LIS(array,n) << endl;}return 0;}

该题复杂度是O(n*n);