Bellovin

E - Bellovin

Time Limit:3000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Peter has a sequence a_1,a_2,...,a_n and he define a function on the sequence -- F(a_1,a_2,...,a_n)=(f_1,f_2,...,f_n), where f_i is the length of the longest increasing subsequence ending with a_i. 

Peter would like to find another sequence b_1,b_2,...,b_n in such a manner that F(a_1,a_2,...,a_n) equals to F(b_1,b_2,...,b_n). Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one. 

The sequence a_1, a_2, ..., a_n is lexicographically smaller than sequence b_1, b_2, ..., b_n, if there is such number i from 1 to n, that a_k = b_k for 1 \le k < i and a_i < b_i.

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 \le n \le 100000) -- the length of the sequence. The second line contains n integers a_1,a_2,...,a_n (1 \le a_i \le 10^9).

Output

For each test case, output n integers b_1,b_2,...,b_n (1 \le b_i \le 10^9) 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

找到每個位置可能的最長遞增子序列（LIS）順次輸出就行了。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
const int maxn = 100010;
int a[maxn],dp[maxn];
int n,t;
int main(){
	scanf("%d",&t);
	while(t--){
		int ac = 0;
		memset(a,0,sizeof(a));
		memset(dp,0,sizeof(dp)); 
		scanf("%d",&n);
		scanf("%d",&ac);
		a[1] = ac;
		int len = 1;
		dp[1] = 1;
		for(int i=2; i<=n; i++){
			scanf("%d",&ac);
			if(a[len]<ac){
				a[++len] = ac;	dp[i] = len;
			}
			else{
				int k = lower_bound(a+1,a+1+len,ac) - a;
				a[k] = ac;
				dp[i] = k;
			}
		}
		for(int i=1; i<=n; i++){
			printf("%d%c",dp[i],i == n?'\n':' ');
		}
	}
	
	return 0;
}