HDU5748-Bellovin（最长上升子序列）

Bellovin

                                                                             Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                                        Total Submission(s): 1219    Accepted Submission(s): 521

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

311055 4 3 2 131 3 5

 

Sample Output

11 1 1 1 11 2 3

 

Source

BestCoder Round #84

#include <iostream>#include <stdio.h>#include <cstring>#include <algorithm>using namespace std;int b[100005],c[100005];int n;int main(){    int t,data;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int num=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&data);            if(i==1) b[++num]=data,c[1]=1;            else if(data>b[num]) b[++num]=data,c[i]=num;            else            {                int k=lower_bound(b+1,b+1+num,data)-b;                b[k]=data;                c[i]=k;            }        }        for(int i=1;i<n;i++)            printf("%d ",c[i]);        printf("%d\n",c[n]);    }    return 0;}