HDU 1423 Greatest Common Increasing Subsequence (LCIS)

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7667    Accepted Submission(s): 2480

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）

垃圾题目没有说输出之间应有一个空行，会PE。

LCIS裸题。

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
int a[555],b[555];
int f[555];
int main()
{
    int T;
    scanf("%d",&T);
    int p=0;
    while(T--)
    {

        int l1,l2;
        scanf("%d",&l1);
        for(int i=1;i<=l1;i++) scanf("%d",&a[i]);
        scanf("%d",&l2);
        for(int i=1;i<=l2;i++) scanf("%d",&b[i]);
        memset(f,0,sizeof f);
        for(int i=1;i<=l1;i++)
        {
            int k=0;//f[k]=0就好
            for(int j=1;j<=l2;j++)
            {
                if(a[i]==b[j])
                {
                    f[j]=max(f[j],f[k]+1);
                }
                else if(a[i]>b[j])
                {
                    if(f[k]<f[j])
                        k=j;
                }
            }
        }
        if(p)
            printf("\n");
        p++;
        int ans=0;
        for(int i=1;i<=l2;i++)
            ans=max(ans,f[i]);
        printf("%d\n",ans);
    }
    return 0;

}