ZOJ2136-Longest Ordered Subsequence

最新推荐文章于 2021-04-01 13:39:47 发布

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最新推荐文章于 2021-04-01 13:39:47 发布

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本文介绍了一种寻找给定数字序列中最长上升子序列长度的高效算法实现。通过使用二分查找方法，该算法能够在O(N log N)的时间复杂度内解决此问题，其中N为序列长度。文章提供了一个C++代码示例，演示了如何处理多个测试用例并输出最长上升子序列的长度。

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Longest Ordered Subsequence

Time Limit: 2 Seconds Memory Limit: 65536 KB

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, the 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 of this sequence 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 contains the length of sequence N (1 <= N <= 1000). The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces.

Output

Output must contain a single integer - the length of the longest ordered subsequence of the given sequence.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1

7
1 7 3 5 9 4 8

Sample Output

Source: Northeastern Europe 2002, Far-Eastern Subregion

题意：给出一串数字，找出最长上升子序列

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while(t--)
    {
        if(cas) printf("\n");
        else cas=1;
        int n,cnt=1;
        scanf("%d",&n);
        int a[1005],b[1005];
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(i==1) b[cnt++]=a[i];
            else
            {
                int k=lower_bound(b+1,b+cnt,a[i])-b;
                if(k==cnt) b[cnt++]=a[i];
                else b[k]=a[i];
            }
        }
        printf("%d\n",cnt-1);
    }
    return 0;
}