HDOJ 5748 Bellovin

Bellovin

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 989    Accepted Submission(s): 442

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

 

Source

BestCoder Round #84

 

Recommend

wange2014

 

這題目貼的我好崩潰= =，直接點連接看題吧……點我點我

這個就是一個很裸的求最長上升子序列，這裏我貼上O(nlogn)的做法。具體分析這裏不貼了。

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
#define N 100000+10
#define INF 0x3f3f3f3f
int a[N],g[N],dp[N];
int main()
{
	int T;
	int i,n;
	scanf("%d",&T);
	while(T--)
	{
		memset(g,INF,sizeof(g));
		scanf("%d",&n);
		for(i=0;i<n;i++)
			scanf("%d",&a[i]),dp[i]=1;
		for(i=0;i<n;i++)
		{
			int k=lower_bound(g+1,g+n+1,a[i])-g;//大於a[i]的最小值的位置 
			g[k]=min(g[k],a[i]); 
			dp[i]=max(k,dp[i]);//在dp[i]處能達到的最長的位置 
		}
		for(i=0;i<n-1;i++)
			printf("%d ",dp[i]);
		printf("%d\n",dp[i]);
	}
	return 0;
}