Gretest Common Increase Subsequence（hdu 1423（LCIS））

/*
http://acm.hdu.edu.cn/showproblem.php?pid=1423
Gretest Common Increasung Subsequence
Time Limit：2000/1000 MS（Java/Others）   メモリLimit：65536/32768 K（Java/Others）
Total Submission(s)：2724   Acceepted Submission(s)：839
Problem Description
This is a problem from ZOJ 2432.To make it yer，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 increase subsequence of both sequences.
 
Sample Input
1
5
1 4 2 5-12
4
-12 1 2 4
 
Sample Output
2
 
ソurce
ACM夏休み合宿チーム練習試合（二）
 
Recommund
lcy
 解析:
 題意と考え方：
 最長の公共上昇シーケンスを求めます。
Acceepted
232 KB
0 ms
C++
805 B
2013-08-03 22:50:40
*/

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include <iostream>
using namespace std;
const int maxn=500+10;
int dp[maxn],a[maxn],b[maxn];//dp[j]  1 j         
int main()
{
	int k,n,i,m,j,T;
	scanf("%d",&T);
	while(T--)
	{    //memset(a,0,sizeof(a));
	     //memset(b,0,sizeof(b));
		scanf("%d",&n);
		for(i=0;i<n;i++)
		scanf("%d",&a[i]);
		scanf("%d",&m);
		for(j=0;j<m;j++)
		scanf("%d",&b[j]);
		memset(dp,0,sizeof(dp));
	for(i=1;i<=n;i++)
	  for(k=0,j=1;j<=m;j++)
	  {
	  	if(b[j-1]<a[i-1]&&dp[j]>dp[k])k=j;
	  	if(b[j-1]==a[i-1]&&dp[k]+1>dp[j])
	  	{dp[j]=dp[k]+1;
	  	 //printf("%d==%d
",j,dp[j]);
	  	}
	  	//printf("%d
",dp[j]);
	  }
	  int mx=0;
	  for(i=0;i<=m;i++)
	   if(dp[i]>mx)
	       mx=dp[i];
	 printf("%d
",mx);
	 if(T)
	 printf("
");
	}
	//system("pause");
return 0;
}