POJ-2533 Longest Ordered Subsequence (线性dp 最长上升子序列)

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 55888 Accepted: 25040

Description

A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence (a₁, a₂, ..., a_N) be any sequence (a_i1, a_i2, ..., a_iK), where 1 <= i₁ < i₂ < ... < i_K <= 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

71 7 3 5 9 4 8

Sample Output

4

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int dp[1001], a[1001];int main(){int n, tot;scanf("%d", &n);for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);}memset(dp, 0, sizeof(dp));dp[1] = a[1];tot = 1;for(int i = 2; i <= n; ++i){if(a[i] > dp[tot]){dp[++tot] = a[i];}else{int pos = lower_bound(dp + 1, dp + 1 + tot, a[i]) - dp;dp[pos] = a[i];}}printf("%d\n", tot);}/*题意:1000个数，求最长上升子序列。思路：线性dp，dp[i]表示长度为i的上升子序列的最后一位的数最小是多少。从左到右扫一遍原数组，维护一下dp值。最终答案就是tot值，复杂度nlog(n)。*/