POJ 2533 Longest Ordered Subsequence(最长路径 dp)

短命女 2024-02-17 19:22 53阅读 0赞

Longest Ordered Subsequence

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 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;2000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;49925</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;22167</td> 
  </tr> 
 </tbody> 
</table>

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.

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

题意很简单，求最长递增子序列的个数；

这题我用的方法是： 未给定起点和终点的最长路解法（DAG上的动态规划），算法复杂度比(nlogn算法 稍微差点，因为hdu1025、1950用的这种方法解，本题就换了种做法)。用到了记忆化搜索 避免超时。

AC代码如下：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=1000+10;
    int d[maxn];
    int v[maxn];
    int n;
    int dp(int i){
    	int& ans=d[i];
    	if(ans>0)return ans;
    	ans=1;
    	for(int j=i+1;j<n;j++){
    		if(v[i]<v[j])
    		   ans=max(ans,dp(j)+1);
    	}
    	return ans;
    }
    int main(){
    	while(scanf("%d",&n)==1){
    		for(int i=0;i<n;i++)
    		 scanf("%d",&v[i]);
    		 memset(d,0,sizeof(d));
    		 int ans=0;
    		 for(int i=0;i<n;i++){
    		 	ans=max(ans,dp(i));
    		 }
    		 printf("%d\n",ans);
    	}
    	return 0;
    }