uva 10887 - Concatenation of Languages

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

开始将字符串合并后用bkdrhash，一直wrong，20位的字符串hash完后超出long long；然后只有分别对2个子串hash，这样后来要比较2个值，最后还是用stl偷懒水过了；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int cmp(const void *a,const void*b)
{return strcmp((char *)a,(char *)b);  }
char s1[1500][11],s2[1500][11],a[2250010][21],s[21];
int t,sum,ans;
int main()
{int T,i,j,k,l1,l2,n,m;
 scanf("%d",&t);
 for (T=1;T<=t;T++)
 {
  scanf("%d%d",&n,&m);
  getchar();
  sum=0;
  for (i=0;i<n;i++)
  gets(s1[i]);
  for (i=0;i<m;i++)
  gets(s2[i]);
  for (i=0;i<n;i++)
  for (j=0;j<m;j++)
  {strcpy(s,s1[i]);
   l1=strlen(s1[i]);
   l2=strlen(s2[j]);
   for (k=l1;k<l1+l2;k++)
   s[k]=s2[j][k-l1];
   s[l1+l2]='\0';
   strcpy(a[sum],s);
   ++sum;
  }
  qsort(a,sum,sizeof(a[0]),cmp);
  k=0; ans=0;
  while (k<sum)
  {++ans; j=k;
   while ((j+1<sum)&&(strcmp(a[j+1],a[k])==0)) ++j;
   k=j+1;
  }
  printf("Case %d: %d\n",T,ans);
 }
 return 0;
}