hdu 1423 Greatest Common Increasing Subsequence（最长公共递增子序列lcis）

http://acm.hdu.edu.cn/showproblem.php?pid=1423

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

 

lcis：dp

 

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<string>
#include<ctime>
#include<queue>
#include<list>
#include<map>
#define INF 9999999
#define MAXN 10000
using namespace std;
//    priority_queue<int,vector<int>,greater<int> > pq;

int dp[550];//b数组到第j个数字时，该数字位置处的公共递增因子数

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n;
        int i,j,a[550],b[550];
        for(i=0;i<n;i++)
            cin>>a[i];
        cin>>m;
        for(i=0;i<m;i++)
            cin>>b[i];
        memset(dp,0,sizeof(dp));
    //    dp[0]=-1;
        for(i=0;i<n;i++)
        {
            int k=0;
            for(j=0;j<m;j++)
            {
                //当前要比较的数值为a[i],所以我们寻找b[j]中比a[i]小，但dp[j]最大的值,找到了就用k记录位置 
                if(a[i]>b[j]&&dp[j]>dp[k])
                    k=j;
                if(a[i]==b[j])
                    dp[j]=dp[k]+1;  //dp[k]值+1就为当前字符的公共递增因子数
            }
        }
        int maks=-1;
        for(i=0;i<m;i++)
            if(maks<dp[i])
                maks=dp[i];
        cout<<maks<<endl;
        if(t)
            cout<<endl;
    }
    return 0;
}

对与测试数据：
1
5
1 4 2 5 -12
4
-12 1 2 4

dp中的数据变化如下：
-12 1 2 4
1 0 1 0 0
4 0 1 0 2
2 0 1 2 2
5 0 1 2 2
-12 1 1 2 2

最后dp中的数据变成了 1 1 2 2