Greatest Common Increasing Subsequence(hdu1423(LCIS)）

/*
http://acm.hdu.edu.cn/showproblem.php?pid=1423
Greatest Common Increasing Subsequence


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2724    Accepted Submission(s): 839




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
 


Source
ACM暑期集训队练习赛（二）
 


Recommend
lcy
 解析：
 题意及思路：
 求最长公共上升序列
Accepted 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]表示1到j中有多少个公共序列
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\n",j,dp[j]);
	  	}
	  	//printf("%d\n",dp[j]);
	  }
	  int mx=0;
	  for(i=0;i<=m;i++)
	   if(dp[i]>mx)
	       mx=dp[i];
	 printf("%d\n",mx);
	 if(T)
	 printf("\n");
	}
	//system("pause");
return 0;
}