HDU-1423-Greatest Common Increasing Subsequence-最长公共上升子序列【模版】

This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

InputEach sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.Outputoutput 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

注意一下控制格式

 #include<stdio.h>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<iomanip>
 #include<string.h>
 using namespace std;

 int a[];
 int b[];
 int dp[];

 int main()
 {
     std::ios::sync_with_stdio(false);
     cin.tie();
     cout.tie();
     int tt;
     cin>>tt;
     while(tt--)
     {
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         memset(dp,,sizeof(dp));
         int p,q;
         cin>>p;
         for(int i=;i<p;i++)
             cin>>a[i];
         cin>>q;
         for(int i=;i<q;i++)
             cin>>b[i];
         int maxx;
         for(int i=;i<p;i++)
         {
             maxx=;//每次都要恢复
             for(int j=;j<q;j++)
             {
                 if(a[i]>b[j])
                     maxx=max(dp[j],maxx);
                 else if(a[i]==b[j])
                     dp[j]=maxx+;
             }
         }
         int ans=;
         for(int i=;i<q;i++)
             ans=max(ans,dp[i]);
         if(tt)
             cout<<ans<<endl<<endl;
         else
             cout<<ans<<endl;
     }
     return ;
 }