Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9961    Accepted Submission(s): 3206


 

Problem Description

This is a problem from ZOJ 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） ​​
 
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代码实现：
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
 
const int MAXN = 1001;
 
int a[MAXN], b[MAXN];
int f[MAXN][MAXN];
int n, m,ans;
 
void init()
{
  memset(f, 0, sizeof(f));
}
void dp()
{
  init();
  int i, j, k;
  for(i = 1; i <= n; i++)
  {
    for(j = 1; j <= m; j++)
    {
      f[i][j] = f[i-1][j]; // if(a[i] != b[j])
      if(a[i] == b[j])
      {
        int MAX = 0;
        for(k = 1; k <= j-1; k++) if(b[j] > b[k]) //枚举最大的f[i-1][k] 
        {
          MAX = max(MAX, f[i-1][k]);
        }
        f[i][j] = MAX+1;
      }
    }
  }
   ans = 0;
  for(int i = 1; i <= m; i++) ans = max(ans, f[n][i]);

}
int main()
{
  int T;
  cin>>T;
    while(T--)
    {
      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]);
      dp();
    
        printf("%d\n", ans);
        if(T)
          printf("\n");
    }
}