hdu-5748 Bellovin(LIS)

题目链接：

Bellovin

Time Limit: 6000/3000 MS (Java/Others)   

 Memory Limit: 131072/131072 K (Java/Others)

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：

 

给一个序列,求出每个位置结尾的最长上升子序列;然后找一个字典序最小的这个函数值相同的子序列;

 

思路：

 

求完这个函数值后这个函数值就是这个字典序最小的序列了;

 

AC代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-8;

int a[N],d[N],g[N];

int main()
{       

        int t;
        read(t);
        while(t--)
        {
            int n;
            read(n);
            For(i,1,n)read(a[i]);
            for(int i=1;i<=n;i++)g[i]=inf;
            for(int i=1;i<=n;i++)
            {
                int k=lower_bound(g+1,g+n+1,a[i])-g;
                d[i]=k;
                g[k]=a[i];
            }
            for(int i=1;i<n;i++)printf("%d ",d[i]);printf("%d
",d[n]);
        }
        return 0;
}