Description
Problem A
Concatenation of Languages
Input File: Standard Input
Output: Standard Output
A language is a set of strings. And the concatenation of two languages is the set of all strings that are formed by concatenating the strings of the second language at the end of the strings of the first language.
For example, if we have two language A and B such that:
A = {cat, dog, mouse}
B = {rat, bat}
The concatenation of A and B would be:
C = {catrat, catbat, dograt, dogbat, mouserat, mousebat}
Given two languages your task is only to count the number of strings in the concatenation of the two languages.
Input
There can be multiple test cases. The first line of the input file contains the number of test cases, T(1≤T≤25). Then T test cases follow. The first line of each test case contains two integers, M and N(M,N<1500), the number of strings in each of the languages. Then the next M lines contain the strings of the first language. The N following lines give you the strings of the second language. You can assume that the strings are formed by lower case letters (‘a’ to ‘z’) only, that they are less than 10 characters long and that each string is presented in one line without any leading or trailing spaces. The strings in the input languages may not be sorted and there will be no duplicate string.
Output
For each of the test cases you need to print one line of output. The output for each test case starts with the serial number of the test case, followed by the number of strings in the concatenation of the second language after the first language.
Sample Input Output for Sample Input
|
2 3 2 cat dog mouse rat bat 1 1 abc cab |
Case 1: 6 Case 2: 1
|
Problem setter: Monirul Hasan
Special Thanks: Shahriar Manzoor
解題思路：
這道題作爲第二道字符串hash題，第一道還是昨晚上的CF的D題，，皮球教給我的，，就是用字符串hash做的，當時我還準備O（n2）暴力枚舉，那樣必定會T，，，
然後呢，今天起來，在VJ上開了個字符串hash專題，這幾天下決心切完。。。
就是字符串的技巧有些問題，再沒什麼問題了，就是把先前有的放進map裏，map<string,int>Map，這個STL，然後掃一遍，hash判重了。
代碼：
# include<cstdio>
# include<iostream>
# include<algorithm>
# include<cstring>
# include<string>
# include<cmath>
# include<queue>
# include<stack>
# include<set>
# include<map>
using namespace std;
# define inf 999999999
# define MAX 1500+4
char A[MAX][14];
char B[MAX][14];
char temp[14<<1];
map<string,int>Map;
int n,m;
void init()
{
Map.clear();
}
void input()
{
cin>>m>>n;
getchar();
for ( int i = 0;i < m;i++ )
{
gets(A[i]);
}
for ( int i = 0;i < n;i++ )
{
gets(B[i]);
}
}
void solve()
{
for ( int i = 0;i < m;i++ )
{
for ( int j = 0;j < n;j++ )
{
strcpy( temp,A[i] );
strcat( temp,B[j] );
if ( !Map[temp] )
{
Map[temp]++;
}
}
}
}
int main(void)
{
int t;cin>>t;
int icase = 1;
while ( t-- )
{
init();
input();
solve();
printf("Case %d: ",icase++);
cout<<Map.size()<<endl;
}
return 0;
}