HDU：5748 Bellovin（LIS+打表）

Bellovin

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1103    Accepted Submission(s): 498

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

 

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题目大意：给你一个数字序列，让你求出以每个a[i]结尾的最长递增子序列长度，并且输出。

解题思路：新开一个dp数组记录以每个a[i]结尾的最长递增子序列长度，其余的都是裸LIS。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int dp[100010];//记录以第i个数字为尾的最长子序列长度 
int d[100010];//辅助数组 
int a[100010];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			d[i]=INF;//初始化辅助数组 
			dp[i]=0;//初始化dp 
		}
		for(int i=1;i<=n;i++)
		{
			int pos=lower_bound(d+1,d+n+1,a[i])-d;//找到当前a[i]能插的点 
			dp[i]=max(dp[i],pos);//更新下dp 
			d[pos]=min(d[pos],a[i]);//将a[i]插入辅助数组 
		}
		for(int i=1;i<n;i++)
		{
			printf("%d ",dp[i]);
		}
		printf("%d\n",dp[n]);
	}
	return 0;
}