UVA - 10887 Concatenation of Languages（hash）

最新推荐文章于 2017-07-18 16:20:00 发布

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本文介绍了一道关于字符串操作的问题，即如何计算两个字符串集合拼接后去重的元素总数，并提供了一个简单的C++实现方案，使用了标准库中的set来确保元素的唯一性。

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

3 2

cat

dog

mouse

rat

bat

1 1

abc

cab

Case 1: 6

Case 2: 1

题目大意：将A集合的词与B集合中的词按要求合成一个词，然后利用哈希表判重，问这个集合一共又几个元素。

解析：直接保存在set中，然后输出set的个数，虽然比较慢，但是代码简单。

注意：空串也算，所以要用gets或者getline输入。

#include <stdio.h>
#include <iostream>
#include <string>
#include <set>
using namespace std;
const int N = 1505;
string fir[N];
string sec[N];
int main() {
	int t;
	int m,n;
	int cas = 1;
	set<string> hash;
	scanf("%d",&t);
	while(t--) {
		scanf("%d%d",&m,&n);
		getchar();
		for(int i = 0; i < m; i++) {
			getline(cin,fir[i]);
		}
		for(int i = 0; i < n; i++) {
			getline(cin,sec[i]);	
		}
		for(int i = 0; i < m; i++) {
			for(int j = 0; j < n; j++) {
				string tmp = fir[i] + sec[j];
				hash.insert(tmp);
			}
		}
		printf("Case %d: %d\n",cas++,(int)hash.size());
		hash.clear();
	}
	return 0;
}