BestCoder Round #84 1002 Bellovin

最新推荐文章于 2024-08-08 09:12:47 发布

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本文介绍了一种高效算法，用于解决给定序列下寻找等长上升子序列的问题。通过使用nlog(n)复杂度的方法替代传统的动态规划，确保了算法在大规模数据集上的可行性。该文详细解释了算法步骤，并提供了实现代码。

Problem Description

Peter has a sequence

a1,a2,...,an and he define a function on the sequence --

F(a1,a2,...,an)=(f1,f2,...,fn), where

fi is the length of the longest increasing subsequence ending with

ai.

Peter would like to find another sequence

b1,b2,...,bn in such a manner that

F(a1,a2,...,an) equals to

F(b1,b2,...,bn). Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one.

The sequence

a1,a2,...,an is lexicographically smaller than sequence

b1,b2,...,bn, if there is such number

i from

1 to

n, that

ak=bk for

1≤k<i and

ai<bi.

Input

There are multiple test cases. The first line of input contains an integer

T, indicating the number of test cases. For each test case:

The first contains an integer

(1≤n≤100000) -- the length of the sequence. The second line contains

n integers

a1,a2,...,an

(1≤ai≤109).

Output

For each test case, output

n integers

b1,b2,...,bn

(1≤bi≤109) denoting the lexicographically smallest sequence.

Sample Input

Sample Output

1
1 1 1 1 1
1 2 3

题意：给一个序列，让换成另一个序列，满足每一位都和原序列的最长上升子序列位数相同，其实就是求每一位的最长上升子序列个数；不能用DP，用nlog（n），不然超时；

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

int dp[100000 + 5];
int b[100000 + 5];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n, temp;
        cin>>n;
        dp[0] = 0;
        b[0] = 0;
        for(int i = 1, j = 0; i <= n; i++)
        {
            cin>>temp;
            if(temp>dp[j])                    //大于最后一位数直接放最后；
            {
                dp[++j] = temp;         
                b[i] = j;
            }
            else
            {
                int add = lower_bound(dp+1,dp+j+1,temp)-dp;          //否则查找到然后替换前面的
                dp[add] = temp;
                b[i] = add;
            }
        }
        cout<<b[1];
        for(int i = 2; i <= n; i++)
            cout<<" "<<b[i];
        cout<<endl;

    }
}