Virus
Time Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
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Description
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Virus
We have a log file, which is a sequence of recorded events. Naturally, the timestamps are strictly increasing.
However, it is infected by a virus, so random records are inserted (but the order of original events is preserved). The backup log file is also infected, but since the virus is making changes randomly, the two logs are now different.
Given the two infected logs, your task is to find the longest possible original log file. Note that there might be duplicated timestamps in an infected log, but the original log file will not have duplicated timestamps.
Input
The first line contains T ( T$ le$100), the number of test cases. Each of the following lines contains two lines, describing the two logs in the same format. Each log starts with an integer n ( 1$ le$n$ le$1000), the number of events in the log, which is
followed by n positive integers not greater than 100,000, the timestamps of the events, in the same order as they appear in the log.
Output
For each test case, print the number of events in the longest possible original log file.
Sample Input
1
9 1 4 2 6 3 8 5 9 1
6 2 7 6 3 5 1
Sample Output
3
最长公共子序列匹配
/*************************************************************************
> File Name: e.cpp
> Author:yuan
> Mail:
> Created Time: 2014年11月09日 星期日 16时37分18秒
************************************************************************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
int a[1007],b[1007],dp[1007];
int t,n,m;
int main()
{
scanf("%d",&t);
while(t--){
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(dp,0,sizeof(dp));
scanf("%d",&n);
for(int i=0;i<n;i++) scanf("%d",&a[i]);
scanf("%d",&m);
for(int i=0;i<m;i++) scanf("%d",&b[i]);
for(int i=0;i<n;i++)
{
int MAX=0;
for(int j=0;j<m;j++)
{
if(a[i]==b[j]){
dp[j]=MAX+1;
}
if(a[i]>b[j]&&MAX<dp[j]){
MAX=dp[j];
}
}
}
int ans=0;
for(int i=0;i<m;i++)
ans=max(ans,dp[i]);
printf("%d
",ans);
}
return 0;
}