题解报告：poj 2533 Longest Ordered Subsequence（LIS）

Posted 2021-01-01 acgoto

Description

A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence ( a₁, a₂, ..., a_N) be any sequence ( a_i1, a_i2, ..., a_iK), where 1 <= i₁ < i₂ < ... < i_K <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8). 
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

7
1 7 3 5 9 4 8

Sample Output

4
解题思路：典型dp：最长上升子序列问题。有两种解法，一种O(n^2)，另一种是O(nlogn)。相关详细的讲解：LIS总结。
AC代码一：朴素O(n^2)算法，数据小直接暴力。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 using namespace std;
 6 const int maxn=1005;
 7 int n,res,dp[maxn],a[maxn];
 8 int main(){
 9     while(~scanf("%d",&n)){
10         memset(dp,0,sizeof(dp));res=0;
11         for(int i=0;i<n;++i)scanf("%d",&a[i]),dp[i]=1;
12         for(int i=0;i<n;++i){
13             for(int j=0;j<i;++j)
14                 if(a[j]<a[i])dp[i]=max(dp[i],dp[j]+1);//只包含i本身长度为1的子序列
15             res=max(res,dp[i]);
16         }
17         printf("%d
",res);
18     }
19     return 0;
20 }

AC代码二：进一步优化，采用二分每次更新最小序列，最终最小序列的长度就是最长上升子序列长度。时间复杂度是O(nlogn)。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 using namespace std;
 6 const int maxn=1005;
 7 const int INF=0x3f3f3f3f;
 8 int n,res,x,dp[maxn];
 9 int main(){
10     while(~scanf("%d",&n)){
11         memset(dp,0x3f,sizeof(dp));
12         for(int i=1;i<=n;++i){
13             scanf("%d",&x);
14             *lower_bound(dp,dp+n,x)=x;//更新最小序列
15         }
16         printf("%d
",lower_bound(dp,dp+n,INF)-dp);
17     }
18     return 0;
19 }