HDU 5748 BestCoder Round #84 Bellovin (LIS)(樹狀陣列)
阿新 • • 發佈:2018-12-24
Bellovin
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 540 Accepted Submission(s): 254
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 a
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 b
The first contains an integer n(1≤n≤100000) -- the length of the sequence. The second line contains n
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 Recommend wange2014 題解:求以ai結尾的最長上升子序列(LIS)為多少。 就是求LIS。 AC程式碼:
//#include<bits/stdc++.h>
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#define N 1000010
using namespace std;
int n,a[N];
int b[N];
int cnt;
int s[N];
int f[N];
int query(int x)
{
int ret =0;
while(x){
ret= max(ret,s[x]);
x-= x&-x;
}
return ret;
}
void update(int x,int y)
{
while(x<=cnt)
{
s[x]=max(s[x],y);
x+=x&-x;
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
cnt=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[++cnt]=a[i];
}
sort(b+1,b+cnt+1);
cnt=unique(b+1,b+cnt+1)-b-1;
for(int i=1;i<=n;i++)
{
a[i]=lower_bound(b+1,b+cnt+1,a[i])-b;
}
for(int i=1;i<=cnt;i++) s[i]=0;
for(int i=1;i<=n;i++)
{
f[i]=query(a[i]-1)+1;
update(a[i],f[i]);
printf("%d%c",f[i],i==n?'\n':' ');
}
}
return 0;
}