美人鱼 hdu 5784

Peter has a sequence a1,a2,...,ana1,a2,...,an and he define a function on the sequence -- F(a1,a2,...,an)=(f1,f2,...,fn)F(a1,a2,...,an)=(f1,f2,...,fn) , where fifi is the length of the longest increasing subsequence ending with aiai .

Peter would like to find another sequence b1,b2,...,bnb1,b2,...,bn

in such a manner that F(a1,a2,...,an)F(a1,a2,...,an)

equals to F(b1,b2,...,bn)F(b1,b2,...,bn)

. Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one.

The sequence a1,a2,...,ana1,a2,...,an

is lexicographically smaller than sequence b1,b2,...,bnb1,b2,...,bn

, if there is such number ii

from 11

to nn

, that ak=bkak=bk

for 1≤k<i1≤k<i

and ai<biai<bi

.InputThere are multiple test cases. The first line of input contains an integer TT

, indicating the number of test cases. For each test case:

The first contains an integer nn

(1≤n≤100000)(1≤n≤100000)

-- the length of the sequence. The second line contains nn

integers a1,a2,...,ana1,a2,...,an

(1≤ai≤109)(1≤ai≤109)

.OutputFor each test case, output nn

integers b1,b2,...,bnb1,b2,...,bn

(1≤bi≤109)(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：

就是求fi，即求以ai为最后一位的最长子序列

解法：

扫描这一个数组a，每一个数进入dp数组，在里面找到<=a[i]的下标最小的指针，在这里面存储a[i]。

lower_bound(dp,dp+n,a[i])-dp+1   这个是找到>=a[i]的最小指针减去dp数组的首指针再加1

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#define inf 0x3f3f3f
using namespace std;
typedef long long ll;
ll a[110000],dp[110000];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%I64d",&a[i]);
        memset(dp,inf,sizeof(dp));
        for(int i=0;i<n-1;i++)
        {
            *lower_bound(dp,dp+n,a[i])=a[i];
            printf("%d ",lower_bound(dp,dp+n,a[i])-dp+1);
        }
        *lower_bound(dp,dp+n,a[n-1])=a[n-1];
        printf("%d\n",lower_bound(dp,dp+n,a[n-1])-dp+1);
    }
    return 0;
}