HDU 5748 BestCoder Round #84 Bellovin （LIS）（树状数组）

Bellovin

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 540    Accepted Submission(s): 254

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

题解：求以ai结尾的最长上升子序列（LIS）为多少。

     就是求LIS。

AC代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#define N 1000010
using namespace std;
int n,a[N];
int b[N];
int cnt;
int s[N];
int f[N];
int query(int x)
{
	int ret =0;
	while(x){
		ret= max(ret,s[x]);
		x-= x&-x;
	}
	return ret;
}
void update(int x,int y)
{
	while(x<=cnt)
	{
		s[x]=max(s[x],y);
		x+=x&-x;
	}
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
    	cnt=0;
    	scanf("%d",&n);
    	for(int i=1;i<=n;i++)
    	{
    		scanf("%d",&a[i]);
    		b[++cnt]=a[i];
		}
		sort(b+1,b+cnt+1);
		cnt=unique(b+1,b+cnt+1)-b-1;
		for(int i=1;i<=n;i++)
		{
			a[i]=lower_bound(b+1,b+cnt+1,a[i])-b;
		}
		for(int i=1;i<=cnt;i++) s[i]=0;
		for(int i=1;i<=n;i++)
		{
			f[i]=query(a[i]-1)+1;
			update(a[i],f[i]);
			printf("%d%c",f[i],i==n?'\n':' ');
		}
	}
    return 0;
}