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2533:Longest Ordered Subsequence(动态规划)

时间:2020-03-29 13:10:54      阅读:53      评论:0      收藏:0      [点我收藏+]

 

总时间限制: 
2000ms
 
内存限制: 
65536kB
描述
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1a2, ..., aN) be any sequence (ai1ai2, ..., 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.
输入
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
输出
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
样例输入
7
1 7 3 5 9 4 8
样例输出
4
来源
Northeastern Europe 2002, Far-Eastern Subregion
memset()函数,不可以初始化int为1,只能初始化为0或-1.

 

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 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main() {//1 7 3 5 9 4 8
 5 
 6     int n,ans=0;
 7     cin>>n;
 8     int a[n],b[n];
 9     memset(b,0,sizeof(b));
10     for(int i=0;i<n;i++){
11         cin>>a[i];
12         for(int j=0;j<i;j++){
13             if(a[i]>a[j]){
14                 b[i]=max(b[i],b[j]+1);
15             }
16         }
17         ans=max(ans,b[i]);
18     }
19     cout<<ans+1<<endl;
20     
21     return 0;
22 }

 

2533:Longest Ordered Subsequence(动态规划)

原文:https://www.cnblogs.com/aiqinger/p/12591628.html

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