ZOJ 3019 Puzzle

解题思路：给出两个数列an,bn,其中an,bn中元素的顺序可以任意改变，求an,bn的LCS

              因为数列中的元素可以按任意顺序排列，所以只需要求出an,bn中的元素有多少个是相同的即可。

反思：一开始以为就是求LCS，一直WA，后来才发现可以按任意顺序排列元素，把相同的元素都排在一起，就是最长的子序列。

Puzzle

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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For sequences of integers a and b, if you can make the two sequences the same by deleting some elements in a and b, we call the remaining sequence "the common sub sequence". And we call the longest one the LCS.

Now you are given two sequences of integers a and b. You can arrange elements in a and b in any order. You are to calculate the max length of the LCS of each arrangement of a and b.

Input

Input will consist of multiple test cases. The first line of each case is two integers N(0 < N < 10000), M(0 < M < 10000) indicating the length of a and b. The second line is N 32-bit signed integers in a. The third line is M 32-bit signed integers in b.

Output

Each case one line. The max length of the LCS of each arrangement of a and b.

Sample Input

5 4
1 2 3 2 1
1 4 2 1

Sample Output

3

#include <stdio.h>  
int a[10010],b[10010];
void sort(int a[],int n)
{
	int i,j,t;
	for(i=0;i<n-1;i++)
	{
		for(j=i+1;j<n;j++)
		{
			if(a[i]>a[j])
			{
				t=a[i];
				a[i]=a[j];
				a[j]=t;
			}
		}
	}
}
int main()
{
	int n,m,i,j,num,k;
	while(scanf("%d %d",&n,&m)!=EOF)
	{
		for(i=0;i<n;i++)
		scanf("%d",&a[i]);
		for(i=0;i<m;i++)
		scanf("%d",&b[i]);
		sort(a,n);
		sort(b,m);
		num=0;
		k=0;
		
		for(i=0;i<n;i++)
		{
			for(j=k;j<m;j++)
			{
				if(a[i]==b[j])
				{
				num++;
				k=j+1;
				break;
			    }
			}
		}
		printf("%d
",num);		
	}
	return 0;
}