1071 Speech Patterns (25 point(s))

文本模式分析与最常用词识别

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本文介绍了一种分析文本中个人偏好词汇的方法，通过识别最常出现的单词来推测说话者的身份，这对于在线验证同一用户是否为真实个体具有重要意义。文章提供了具体的输入输出规范及示例代码，展示了如何使用C++的STL容器和迭代器进行高效处理。

1071 Speech Patterns (25 point(s))

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when validating, for example, whether it's still the same person behind an online avatar.

Now given a paragraph of text sampled from someone's speech, can you find the person's most commonly used word?

Input Specification:

Each input file contains one test case. For each case, there is one line of text no more than 1048576 characters in length, terminated by a carriage return \n. The input contains at least one alphanumerical character, i.e., one character from the set [0-9 A-Z a-z].

Output Specification:

For each test case, print in one line the most commonly occurring word in the input text, followed by a space and the number of times it has occurred in the input. If there are more than one such words, print the lexicographically smallest one. The word should be printed in all lower case. Here a "word" is defined as a continuous sequence of alphanumerical characters separated by non-alphanumerical characters or the line beginning/end.

Note that words are case insensitive.

Sample Input:

Can1: "Can a can can a can?  It can!"

Sample Output:

can 5

字符串处理、STL的map使用。

1. 熟悉STL各种容器的迭代器的使用，常常会事半功倍。

2. 注意有可能遍历到最后，word如果不为空，则需要把这个单词计入。

#include<iostream>
#include<cstring>
#include<ctype.h>
#include<map>
using namespace std;
int main(void){
	string s;
	getline(cin,s);
	string word = "";
	map<string,int> mp;
	for(int i=0;i<s.length();i++){
		if(isalnum(s[i])){
			word+=tolower(s[i]);
		}else{
			if(word.length()>0){
				mp[word]++;
				word = "";
			}
		}
	}
	if(word.length()>0) mp[word]++;
	int max = 0;string ans;
	map<string,int>::iterator it;
	for(it=mp.begin();it!=mp.end();it++){
		if(it->second>max){
			max = it->second;
			ans = it->first;
		}
	}
	cout<<ans<<" "<<max<<endl;
	return 0;
}