HDU Problem 5748 Bellovin 【LIS】

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本篇介绍了一个算法问题：给定序列An，如何找到一个序列Bn，使得Bn的最长递增子序列长度等于An对应位置的值，并且所有可能的Bn中字典序最小。

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

(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

Sample Output

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;
}