hdu 5748 Bellovin【nlogn最长递增子序列】

Bellovin

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 354    Accepted Submission(s): 178

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 integerT, 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 

中文题意：

思路：

1、显然，其实令bi为最小字典序，其实就是在求f1，f2，f3，......................

2、又显然，n^2会超时，那么我们用nlogn的方式来解决。

Ac代码：

#include <iostream>#include<stdio.h>using namespace std;int a[1000000],b[1000000],c[1000000];int find(int *a,int len,int n)//若返回值为x,则a[x]>=n>a[x-1]{    int left=0,right=len,mid=(left+right)/2;    while(left<=right)    {        if(n>a[mid]) left=mid+1;        else if(n<a[mid]) right=mid-1;        else return mid;        mid=(left+right)/2;    }    return left;}void fill(int *a,int n){    for(int i=0;i<=n;i++)        a[i]=1000000010;}int main(){    int max,i,j,n;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        fill(c,n+1);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        c[0]=-1;//     …………………………………1        c[1]=a[0];//         …………………………2        b[0]=1;//      …………………………………3        for(i=1;i<n;i++)//           ………………4        {            j=find(c,n+1,a[i]);//  …………………5            c[j]=a[i];// ………………………………6            b[i]=j;//……………………………………7        }        for(int i=0;i<n-1;i++)        {            printf("%d ",b[i]);        }        printf("%d\n",b[n-1]);    }}