POJ 2533 Longest Ordered Subsequence-白红宇

POJ 2533 Longest Ordered Subsequence

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发布时间：2019-06-27

本文共 1718 字，大约阅读时间需要 5 分钟。

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_{i_K}), 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 分析 最长上升子序列。定义d[i]为以第i个元素为起始的LIS，那么d[i]=max(d[j])+1,(j>i)。初始是d[n]=1。

#include
     
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                using namespace std;typedef long long LL;const int maxn = 1e4+5;const int mod = 77200211+233;typedef pair
                
                  pii;#define X first#define Y second#define pb push_back//#define mp make_pair#define ms(a,b) memset(a,b,sizeof(a))const int inf = 0x3f3f3f3f;#define lson l,m,2*rt#define rson m+1,r,2*rt+1typedef long long ll;#define N 100010int a[maxn];int dp[maxn];int main(){ int n; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); ms(dp,-1); for(int i=n;i>=1;i--){ dp[i]=1; int res=0; for(int j=i+1;j<=n;j++){ if(a[i]

转载于:https://www.cnblogs.com/fht-litost/p/8855310.html

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