hdu 1423 Greatest Common Increasing Subsequence 最大公共上升子序列 DP

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4590    Accepted Submission(s): 1462

Problem Description

This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

 

Input

Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

 

Output

output print L - the length of the greatest common increasing subsequence of both sequences.

 

Sample Input

   
   
   
   

    
    
    
    1

5
1 4 2 5 -12
4
-12 1 2 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2
   
   
   
   

 

最近一直在做DP题。。现在的进步是，可以看得懂别人的题解了，。

上一个百度文库的文章讲的还是非常不错的。 最长公共上升子序列(LCIS)的平方算法

下面是用O(n^3)的算法。还会更新O(n^2)的算法 。

#include <stdio.h>
#include <string.h>
#define MAX 600

int dp[MAX][MAX] ;
int max(int a ,int b)
{
	return a>b?a:b ;
}

int main()
{
	int num1[MAX],num2[MAX], t;
	scanf("%d",&t);
	while(t--)
	{
		int n , m ;
		scanf("%d",&n);
		for(int i = 1 ; i <= n ; ++i)	scanf("%d",&num1[i]);
		scanf("%d",&m);
		for(int i = 1 ; i <= m ; ++i)	scanf("%d",&num2[i]);
		memset(dp,0,sizeof(dp)) ;
		int ans  = 0 ;
		for(int i = 1 ; i <= n ; ++i)
		{
			for(int j = 1 ; j <= m ; ++j)
			{
				if(num1[i]!=num2[j])	dp[i][j] = dp[i-1][j];
				if(num1[i] == num2[j])
				{
					dp[i][j] = 1 ;
					for(int k = 1 ; k < j ; ++k)
					{
						if(num2[j]>num2[k])	dp[i][j] = max(dp[i][j],dp[i-1][k]+1) ;
					}
				}
				
				ans = max(ans,dp[i][j]) ;
			}
		}
		printf("%d\n",ans) ;
		if(t != 0)
		{
			printf("\n") ;
		}
	}
	
	return 0 ;
}

-----------------------------------------------------------------分割线------------------------------------------------------------

晚上更新：

捣鼓了O(n^2)+一维数组的算法，，不得不佩服搞出这个算法的人，神牛啊！！我看都这么费劲，人家却能发明出来。哎，深刻的感受到了这个世界恶意

o(╯□╰)o

看代码：

#include <stdio.h>
#include <string.h>
#define MAX 505 
int getMax(int a , int b)
{
	return a>b?a:b ;
}

int main()
{
	int num1[MAX],num2[MAX],dp[MAX],t;
	scanf("%d",&t);
	while(t--)
	{
		int n , m ;
		scanf("%d",&n);
		for(int i = 1 ; i <= n ; ++i)	scanf("%d",&num1[i]);
		scanf("%d",&m);
		for(int i = 1 ; i <= m ; ++i)	scanf("%d",&num2[i]);
		memset(dp,0,sizeof(dp)) ;
		int ans  = 0 ;
		for(int i = 1 ; i <= n ; ++i)
		{
			int max = 0 ;
			for(int j = 1 ; j <= m ; ++j)
			{
				if(num1[i]>num2[j] && max < dp[j])	max = dp[j] ;
				if(num1[i] == num2[j])
				{
					dp[j] = max+1 ;
				}
				ans = getMax(ans,dp[j]) ;
			}
		
		}
		printf("%d\n",ans) ;
		if(t!=0)
		{
			printf("\n") ;
		}
	}
	return 0 ;
}