hdu 5748 Bellovin（LIS）

Bellovin

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 162    Accepted Submission(s): 96

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 n (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

3
1
10
5
5 4 3 2 1
3
1 3 5

 

Sample Output

1
1 1 1 1 1
1 2 3

 

题意：给n个数字表示a序列，求字典序最小的b序列中每个位置的LIS跟a序列一样

思路：字典序最小其实就是a序列的LIS

所以此题只要求出a序列中1~n的LIS值即可

用O（nlogn）的算法可以得到

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 100050
int a[N],n;
int ans[N],pre[N],d[N];
int main()
{
    int T,t;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        ans[1]=a[1];
        int len=1;
        d[0]=0;
        d[1]=1;
        for(int i=2; i<=n; i++)
        {
            if(a[i]>ans[len])
            {
                ans[++len]=a[i];
                d[i]=len;
            }
            else
            {
                int pos=lower_bound(ans+1,ans+1+len,a[i])-ans;///注意这个地方是减ans，长度要从1开始
                ans[pos]=a[i];
                d[i]=pos;
            }
        }
        for(int i=1; i<n; i++)
            printf("%d ",d[i]);
        printf("%d\n",d[n]);
    }
    return 0;
}