Judging

Time Limit: 20 Sec Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=114147

Description

The NWERC organisers have decided that they want to improve the automatic grading of the submissions for the contest, so they now use two systems: DOMjudge and Kattis. Each submission is judged by both systems and the grading results are compared to make sure that the systems agree. However, something went wrong in setting up the connection between the systems, and now the jury only knows all results of both systems, but not which result belongs to which submission! You are therefore asked to help them figure out how many results could have been consistent.

Input

The input consists of:

one line with one integer n (1≤n≤105), the number of submissions;

n lines, each with a result of the judging by DOMjudge, in arbitrary order;

n lines, each with a result of the judging by Kattis, in arbitrary order.

Each result is a string of length between 5 and 15 characters (inclusive) consisting of lowercase letters.

Output

Output one line with the maximum number of judging results that could have been the same for both systems.

Sample Input

correct

wronganswer

correct

timelimit

wronganswer

correct

timelimit

correct

timelimit

Sample Output

HINT

题意

有两个机器，每个机器都会返回n个串，然后问你有多少个串是在两个地方都出现过

题解：

双hash一下，然后用map存一下，然后搞一搞就好了……

代码：

//qscqesze#include 
     
      #include 
      
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                   typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 200001#define mod 10007#define eps 1e-9int Num;char CH[20];//const int inf=0x7fffffff; //нчоч╢Сconst int inf=0x3f3f3f3f;/*inline void P(int x){ Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts("");}*/inline ll read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}inline void P(int x){ Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts("");}//**************************************************************************************int get_hash(char *key){ ll N=1998585857; long long h=0; while(*key) h=(h*127+(*key++)+N)%N; return h%N;}int get_hash2(char *key){ ll N=127398127; long long h=0; while(*key) h=(h*127+(*key++)+N)%N; return h%N;}char s[20];map< pair
                   
                     ,int>H;int main(){ int n=read(); for(int i=0;i
                    
                      a; a.first=get_hash(s); a.second=get_hash2(s); H[a]++; } int ans=0; for(int i=0;i
                     
                       a; a.first=get_hash(s); a.second=get_hash2(s); if(H[a]) { ans++; H[a]--; } } printf("%d\n",ans);}

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