HDU-5748-Bellovin【LIS】

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Bellovin

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 996    Accepted Submission(s): 447

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 1k<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 (1n100000) -- the length of the sequence. The second line contains n integers a1,a2,...,an (1ai109).
Output
For each test case, output n integers b1,b2,...,bn (1bi109) denoting the lexicographically smallest sequence.
Sample Input
311055 4 3 2 131 3 5

Sample Output
11 1 1 1 11 2 3

题意:

问题描述
Peter有一个序列a_1,a_2,...,a_na1,a2,...,an. 定义F(a_1,a_2,...,a_n)=(f_1,f_2,...,f_n)F(a1,a2,...,an)=(f1,f2,...,fn), 其中f_ifi是以a_iai结尾的最长上升子序列的长度.Peter想要找到另一个序列b_1,b_2,...,b_nb1,b2,...,bn使得F(a_1,a_2,...,a_n)F(a1,a2,...,an)F(b_1,b_2,...,b_n)F(b1,b2,...,bn)相同. 对于所有可行的正整数序列, Peter想要那个字典序最小的序列.序列a_1, a_2, ..., a_na1,a2,...,anb_1, b_2, ..., b_nb1,b2,...,bn字典序小, 当且仅当存在一个正整数ii (1 \le i \le n)(1in)满足对于所有的kk (1 \le k < i)(1k<i)都有a_k = b_kak=bk并且a_i < b_iai<bi.
输入描述
输入包含多组数据, 第一行包含一个整数TT表示测试数据组数. 对于每组数据:第一行包含一个整数nn (1 \le n \le 100000)(1n100000)表示序列的长度. 第二行包含nn个整数a_1,a_2,...,a_na1,a2,...,an (1 \le a_i \le 10^9)(1ai109).
输出描述
对于每组数据, 输出nn个整数b_1,b_2,...,b_nb1,b2,...,bn (1 \le b_i \le 10^9)(1bi109)表示那个字典序最小的序列.
#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int MAX=1e5+10;const int INF=0x3f3f3f3f;int n;int dp[MAX];int main(){int t;scanf("%d",&t);while(t--){scanf("%d",&n);fill(dp,dp+n,INF);int a,len;for(int i=0;i<n;i++){scanf("%d",&a);len=lower_bound(dp,dp+n,a)-dp;dp[len]=a; // 把数插入数组printf(i==0?"%d":" %d",len+1);}puts("");}return 0;}

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