Bellovin
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1008 Accepted Submission(s): 452
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.
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).
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
Recommend
题意:给一个序列An, 让你找出An元素的每个位置的LIS最小的字典序序列Bn。
An的LIS就是字典序最小的
#include <bits/stdc++.h> using namespace std; const int MAXN = 100010; const int INF = 1e9; int ar[MAXN], g[MAXN], dp[MAXN]; int main() { int t; scanf("%d", &t); while (t--) { int n; scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", &ar[i]); g[i] = INF; } for (int i = 1; i <= n; i++) { int k = lower_bound(g + 1, g + 1 + n, ar[i]) - g; dp[i] = k; g[k] = min(g[k], ar[i]); } for (int i = 1; i <= n; i++) { if (i == n) printf("%d ", dp[i]); else printf("%d ", dp[i]); } } return 0; }