题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5748

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

Source

BestCoder Round #84

题目大意：

Peter有一个序列a_1,a_2,...,a_na?1??,a?2??,...,a?n??. 定义F(a_1,a_2,...,a_n)=(f_1,f_2,...,f_n)F(a?1??,a?2??,...,a?n??)=(f?1??,f?2??,...,f?n??), 其中f_if?i??是以a_ia?i??结尾的最长上升子序列的长度.

Peter想要找到另一个序列b_1,b_2,...,b_nb?1??,b?2??,...,b?n??使得F(a_1,a_2,...,a_n)F(a?1??,a?2??,...,a?n??)和F(b_1,b_2,...,b_n)F(b?1??,b?2??,...,b?n??)相同. 对于所有可行的正整数序列, Peter想要那个字典序最小的序列.

序列a_1, a_2, ..., a_na?1??,a?2??,...,a?n??比b_1, b_2, ..., b_nb?1??,b?2??,...,b?n??字典序小, 当且仅当存在一个正整数ii (1 \le i \le n)(1≤i≤n)满足对于所有的kk (1 \le k < i)(1≤k<i)都有a_k = b_ka?k??=b?k??并且a_i < b_ia?i??<b?i??.

解题思路：

定义一个l数组，初始化都为最大值。再循环输入的数。找到l数组中第一个比a大的数，记录下标即可。

详细过程如下图。

详见代码。

（第一种方法：通过二分的方法来找第一个比a大的数的下标值）

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

#define INF (1<<31-1)

int l[100010];

int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int n,a;
        scanf("%d",&n);
        for(int i=0; i<=n; i++)
            l[i]=INF;
        for (int i=0; i<n; i++)
        {
            scanf("%d",&a);
            int j;
            int ll,r;
            ll=0,r=n;
            while(ll<=r)
            {

                int mid=(r+ll)/2;
                if (a<l[mid])
                    r=mid-1;
                else if (a>l[mid])
                    ll=mid+1;
                else
                {
                    ll=mid;
                    break;
                }

            }
            printf (i==0?"%d":" %d",ll+1);
            l[ll]=a;
        }
        printf("\n");
    }
    return 0;
}

第二种方法（通过库的函数实现）

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

#define INF (1<<31-1)

int l[100010];

int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int n,a;
        scanf("%d",&n);
        for(int i=0; i<=n; i++)
            l[i]=INF;
        for (int i=0; i<n; i++)
        {
            scanf("%d",&a);
            int ll=lower_bound(l,l+n,a)-l;//得到l数组里第一个比a大的数
            printf (i==0?"%d":" %d",ll+1);
            l[ll]=a;
        }
        printf("\n");
    }
    return 0;
}