hdu 5748(LIS)

Bellovin

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 302    Accepted Submission(s): 148

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

 

题意:求以ai（0<i<n）结尾的最长上升子序列的个数.

题解:在LIS的O（n*log(n)）的模板里面找到加一个dp数组记录每一位的最长上升子序列个数，顺序输出即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
using namespace std;
const int N = 100005;
int a[N];
int dp[N],c[N];

int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        int MAX = 1,j=1;
        c[1] = a[1];
        dp[1] = 1;
        for(int i=2; i<=n; i++)
        {
            if(a[i]<=c[1]) j = 1;
            else if(a[i]>c[MAX]) j = ++MAX;
            else j = lower_bound(c+1,c+1+MAX,a[i])-c;
            c[j] = a[i];
            dp[i] = j;
        }
        for(int i=1; i<n; i++)
        {
            printf("%d ",dp[i]);
        }
        printf("%d
",dp[n]);
    }
    return 0;
}