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美人鱼 hdu 5784

时间:2019-01-02 17:24:15      阅读:154      评论:0      收藏:0      [点我收藏+]

标签:sample   ring   eof   multi   lex   row   iostream   tip   put   

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 1k<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 (1n100000)(1≤n≤100000) -- the length of the sequence. The second line contains nn integers a1,a2,...,ana1,a2,...,an (1ai109)(1≤ai≤109) .
OutputFor each test case, output nn integers b1,b2,...,bnb1,b2,...,bn (1bi109)(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;
}

  

美人鱼 hdu 5784

标签:sample   ring   eof   multi   lex   row   iostream   tip   put   

原文地址:https://www.cnblogs.com/EchoZQN/p/10209335.html

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