poj-2533-LIS模板题

LongestOrderedSubsequence TimeLimit: 2000MS   MemoryLimit: 65536K TotalSubmissions: 53144   Accepted: 23696 Description Anumericsequenceof ai isorderedif a1 < a2 <...< aN.Letthesubsequenceofthegivennumericsequence(a1, a2,..., aN) beanysequence(ai1, ai2,..., aiK),where1<= i1 < i2 <...< iK <= N.Forexample,sequence (1,7,3,5,9,4,8)hasorderedsubsequences,e.g.,(1,7),(3,4,8)andmanyothers.Alllongestorderedsubsequencesareoflength4,e.g.,(1,3,5,8). Yourprogram,whengiventhenumericsequence,mustfindthelengthofitslongestorderedsubsequence. Input ThefirstlineofinputfilecontainsthelengthofsequenceN.Thesecondlinecontainstheelementsofsequence-Nintegersintherangefrom0to10000each,separatedbyspaces.1<=N<=1000 Output Outputfilemustcontainasingleinteger-thelengthofthelongestorderedsubsequenceofthegivensequence. SampleInput 7 1735948 SampleOutput 4 Source NortheasternEurope2002,Far-EasternSubregion

方法一：O(n2)法

#include<iostream>
#include<algorithm>
usingnamespacestd;
#defineMAX100000

intn,A[MAX+1],L[MAX+1];//A[]含有输入数据，L[i]代表以i结尾的递增序列的长度
intlis()
{
intres=1;
for(inti=0;i<n;i++)//全部初始化为1
L[i]=1;
for(inti=1;i<n;i++)
{
for(intj=0;j<i;j++)
{
if(A[i]>A[j]&&L[i]<L[j]+1)//如果前面的数字小，且其对应长度没有前面那个数字对应长度长
L[i]=L[j]+1;
}
res=max(res,L[i]);
}
returnres;
}
intmain()
{
cin>>n;
for(inti=0;i<n;i++)
{
cin>>A[i];
}
cout<<lis()<<endl;
return0;
}方法二：O(nlogn)法

#include<iostream>
#include<algorithm>
usingnamespacestd;
#defineMAX100000

intn,A[MAX+1],L[MAX+1];
intlis()
{
L[0]=A[0];
intlength=1;
for(inti=1;i<n;i++)
{
if(L[length-1]<A[i])//如果末尾元素比当前元素小，序列加长
L[length++]=A[i];
else			//如果末尾元素大，则序列中按大小插入替换当前元素，序列长度不变
*lower_bound(L,L+length,A[i])=A[i];
}
returnlength;
}
intmain()
{
cin>>n;
for(inti=0;i<n;i++)
{
cin>>A[i];
}
cout<<lis()<<endl;
return0;
}