1071 Speech Patterns (25)

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

题目大意：给出一行字符串，要求统计出现最多的单词的个数。如果出现字数相同，输出字典序最小的那个。单词的定义为：大小写字母+数字的组合，如果碰到空格或者符号，则视为分隔。单词对大小写不敏感，输出时全部转为小写。

分析：使用map建立字符串与数字的对应关系。由于map是自排序的，统计完单词之后，找到第一个出现次数最多的字符串即可。

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
    //freopen("in.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    map<string,int>mp;
    string ss;
    while(cin>>ss&&ss!="\n")
    {
        string temp;
        for(int i=0;ss[i];++i)
        {
//            db2(i,ss[i]);
            if(ss[i]>='A'&&ss[i]<='Z')ss[i]=tolower(ss[i]),temp+=ss[i];
            else if((ss[i]>='a'&&ss[i]<='z')||(ss[i]>='0'&&ss[i]<='9'))temp+=ss[i];
            else if(!temp.empty())
            {
//                db2(ss,temp);
                if(mp.find(temp)==mp.end())mp[temp]=1;
                else mp[temp]++;
                temp.clear();
            }
        }
        if(!temp.empty())
            if(mp.find(temp)==mp.end())mp[temp]=1;
            else mp[temp]++;
    }

//    db1(mp.size());
//    for(auto it=mp.begin();it!=mp.end();++it)
//        cout<<it->first<<' '<<it->second<<endl;
//    cout<<endl;

    map<string,int>::iterator it=mp.begin();
    string temp=it->first;
    int cnt=it->second;
    for(++it;it!=mp.end();++it)
    {
        if(it->second>cnt)cnt=it->second,temp=it->first;
    }
    cout<<temp<<' '<<cnt<<endl;


    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}