J - Greatest Common Increasing Subsequence HDU - 1423

最新推荐文章于 2019-07-16 19:32:19 发布

原创最新推荐文章于 2019-07-16 19:32:19 发布 · 152 阅读

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本文介绍了解决ZOJ2432问题的方法，该问题是求两个序列的最大公共递增子序列。通过动态规划算法，结合最长公共子序列与最长递增子序列的概念，有效地解决了这一问题。

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

题意：求最大公共递增子数列。

题解：这是一道最长公共子序列与最长递增子序列的综合题，我么们应该用动态规划求出最长公共子序列的同时调用函数（看看是否是递增的。

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
using namespace std;
int a[550],f[550][550],b[550];
int find(int i,int j)
{
    int temp=0;
    for(int k1=1; k1<i; k1++)
    {
        for(int k2=1; k2<j; k2++)
        {
            if(a[k1]<a[i]&&b[k2]<b[j])
            {
                if(f[k1][k2]>temp) temp=f[k1][k2];
            }
        }
    }
    return temp;
}
int main()
{
    int T;
    cin>>T;
    for(int t=1; t<=T; t++)
    {
        int n,m,maximum=-1;
        memset(f,0,sizeof(f));
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        scanf("%d",&m);
        for(int i=1; i<=m; i++)
        {
            scanf("%d",&b[i]);
        }
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                if(a[i]==b[j])
                {
                    f[i][j]=find(i,j)+1;
                }
                else
                {
                    f[i][j]=max(find(i-1,j),find(i,j-1));
                }
                if(f[i][j]>maximum) maximum=f[i][j];
            }
        }
        cout<<maximum<<endl;
        if(t!=T) cout<<endl;
    }
    return  0;
}