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

311055 4 3 2 131 3 5

 

Sample Output

11 1 1 1 11 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 0x3f3f3f3fint 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;}