1071. Speech Patterns (25)

题目：

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

注意：

1、大概是受前面那些题目的影响，一看到这里这么大的数据量，以为又需要先将string转化成int然后hash索引等等，结果发现想多了。

2、这里直接用map来建立索引表就好了，处理好每个charactor的开始和截断，每次得到一个新输入的单词就查询map中是否已经存在，若存在则count+1，不存在则insert。

3、注意printf不能直接输出string类型的字符串，这里还是用cout输出吧。

代码：

//1071
#include<iostream>
#include<vector>
#include<map>
#include<string>
using namespace std;

int main()
{
	vector<int>count;
	vector<string>charactor;
	map<string,int>str2index;
	char t;
	string c;
	while(1)
	{
		scanf("%c",&t);
		if(t>='A'&&t<='Z')
			c += t-'A'+'a';
		else if((t>='a'&&t<='z')||(t>='0'&&t<='9'))
			c += t;
		else if(!c.empty())
		{//if c is not empty, then this charactor has been ended and count it
			map<string,int>::iterator iter=str2index.find(c);
			if(iter==str2index.end())
			{//if c is not existed in map, insert into the map
				str2index.insert(pair<string,int>(c,charactor.size()));
				charactor.push_back(c);
				count.push_back(1);
			}
			else
			{//if c is existed in map, add its count
				++count[iter->second];
			}
			c.clear();
		}
		if(t=='\n')
			break;
	}
	int index=0,times=count[0];
	for(int i=1;i<count.size();++i)
	{//find the maximum count
		if(count[i]>times)
		{
			index=i;
			times=count[i];
		}
	}
	cout<<charactor[index]<<' '<<times;
	return 0;
}