首页 > 代码库 > POJ 2533 - Longest Ordered Subsequence（最长上升子序列） 题解
POJ 2533 - Longest Ordered Subsequence（最长上升子序列） 题解
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题目链接：http://poj.org/problem?id=2533
Description
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= 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.
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
71 7 3 5 9 4 8
Sample Output
4
Source
Northeastern Europe 2002, Far-Eastern Subregion
分析：
DP不解释
AC代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 int ans,n,num[1002],line[1002]; 7 //line[i]的意义是已经算到第i个数的最长公共子序列 8 int main() 9 {10 scanf("%d",&n);11 for(int i = 1;i <= n;++ i)12 scanf("%d",&num[i]),line[i] = 1;13 for(int i = 2;i <= n;++ i)14 for(int j = 1;j < i;++ j)15 if(num[j] < num[i])16 line[i] = max(line[i],line[j]+1);17 for(int i = 1;i <= n;++ i)18 ans = max(ans,line[i]);19 printf("%d\n",ans);20 return 0;21 }
POJ 2533 - Longest Ordered Subsequence（最长上升子序列） 题解
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