HDOJ 5748 Bellovin(LIS)

Bellovin

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 816    Accepted Submission(s): 372

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 integerT, 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, outputn 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

 

上周BC靠这一题上了1700，记录一下吧。

中文题面：点击打开链接  

题解：我们要找到的以ai为最后一个字符的当前段严格递增的最长上升子序列的长度，就是要判断ai前面有多少个比它小的数。且O(n^2)算法超时，所以我用了O(n*logn)的算法。

代码如下:

#include<cstdio>#include<cstring>#include<algorithm>#define maxn 100010#define INF 1000000100using namespace std;int f[maxn];int b[maxn];int dp[maxn];int main(){int t,n,i;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;++i)scanf("%d",&b[i]);fill(dp,dp+n,INF);for(i=0;i<n;++i){*lower_bound(dp,dp+n,b[i])=b[i];//往序列中第一个>=b[i]的地方插入b[i] f[i]=upper_bound(dp,dp+n,b[i])-dp;//找到序列中第一个大于b[i]的位置 }int cnt=1;for(i=0;i<n;++i){if(i!=n-1)printf("%d ",f[i]);elseprintf("%d\n",f[i]);}}return 0;}