HDU Problem 5748 Bellovin 【LIS】

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原文链接：http://www.cnblogs.com/cniwoq/p/6770834.html

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本文介绍了一种算法，用于解决给定序列An，找到一个序列Bn，使得An每个位置的最长递增子序列长度等于Bn对应位置的最长递增子序列长度，并确保Bn为字典序最小。

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.

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\n", dp[i]);
            else printf("%d ", dp[i]);
        }
    }
    return 0;
}