HDOJ Greatest Common Increasing Subsequence（LCIS最长公共上升子序列）_the real incantation is their common increasing su-优快云博客

本文链接：https://blog.youkuaiyun.com/Dreamlandz/article/details/52662550

本文介绍了一种求解最长公共上升子序列(LCIS)的算法实现，并提供了一个具体的示例程序。该算法适用于两个整数序列，通过动态规划找出它们之间的最长公共上升子序列的长度。

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Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6999 Accepted Submission(s): 2279

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

思路：就是求最长公共上升子序列。

代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
int t,n,m;
int a[501],b[501];
int dp[501][501];
int LCIS()
{
    int i,j;
    int max;
    for(i=1;i<=n;i++)
    {
        max=0;
        for(j=1;j<=m;j++)
        {
            dp[i][j]=dp[i-1][j];
            if(a[i]>b[j]&&max<dp[i-1][j])
            {
                max=dp[i-1][j];
            }
            if(a[i]==b[j])
            {
                dp[i][j]=max+1;
            }
        }
    }
    max=0;
    for(i=1;i<=m;i++)
    {
        if(max<dp[n][i])
        {
            max=dp[n][i];
        }
    }
    return max;
}
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        scanf("%d",&m);
        for(i=1;i<=m;i++)
        {
            scanf("%d",&b[i]);
        }
        memset(dp,0,sizeof(dp));
        printf("%d\n",LCIS());
        if (t)
            printf("\n");
    }
    return 0;
}

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