程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> SWERC13 Trending Topic

SWERC13 Trending Topic

編輯：C++入門知識

SWERC13 Trending Topic

map暴力。。。

Imagine you are in the hiring process for a company whose principal activity is the analysis
of information in the Web. One of the tests consists in writing a program for maintaining up to
date a set of trending topics. You will be hired depending on the efficiency of your solution.
They provide you with text from the most active blogs. The text is organised daily and you
have to provide the sorted list of the N most frequent words during the last 7 days, when asked.
INPUT
Each input file contains one test case. The text corresponding to a day is delimited by tag
. Queries of top N words can appear between texts corresponding to two different days.
A top N query appears as a tag like . In order to facilitate you the process of reading
from input, the number always will be delimited by white spaces, as in the sample.
Notes:
? All words are composed only of lowercase letters of size at most 20.
? The maximum number of different words that can appear is 20000.
? The maximum number of words per day is 20000.
? Words of length less than four characters are considered of no interest.
? The number of days will be at most 1000.
? 1 ≤ N ≤ 20
OUTPUT
The list of N most frequent words during the last 7 days must be shown given a query. Words
must appear in decreasing order of frequency and in alphabetical order when equal frequency.
There must be shown all words whose counter of appearances is equal to the word
at position N. Even if the amount of words to be shown exceeds N.

SAMPLE INPUT

imagine you are in the hiring process of a company whose
main business is analyzing the information that appears
in the web

a simple test consists in writing a program for
maintaining up to date a set of trending topics

you will be hired depending on the efficiency of your solution

they provide you with a file containing the text
corresponding to a highly active blog

the text is organized daily and you have to provide the
sorted list of the n most frequent words during last week
when asked

each input file contains one test case the text corresponding
to a day is delimited by tag text

the query of top n words can appear between texts corresponding
to two different days

blah blah blah blah blah blah blah blah blah
please please please

2
Problem IProblem I
Trending Topic
SAMPLE OUTPUT

analyzing 1
appears 1
business 1
company 1
consists 1
date 1
depending 1
efficiency 1
hired 1
hiring 1
imagine 1
information 1
main 1
maintaining 1
process 1
program 1
simple 1
solution 1
test 1
that 1
topics 1
trending 1
whose 1
will 1
writing 1
your 1

text 4
corresponding 3
file 2
provide 2
test 2
words 2

blah 9
text 4
corresponding 3
please 3

#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

typedef pair pII;

map Hash;
vector dy[11];
string rHash[20200];
int day_sum[11][20200];
char cache[30];
int now=9,pre=0,id=1;
int arr[20020],na;
string rss[20020];
bool vis[20020];

void DEBUG(int x)
{
    int sz=dy[x].size();
    for(int i=0;ib.times;
    else
        return a.word= sig &&vis[dy[now][i]]==false)
        {
            rsp[rn++]=(RSP){times,rHash[dy[now][i]]};
            vis[dy[now][i]]=true;
        }
    }
    sort(rsp,rsp+rn,cmpRSP);
    printf("\n",k);
    for(int i=0;i\n");
}

int main()
{
    while(scanf("%s",cache)!=EOF)
    {
        if(strcmp(cache,"")==0)
        {
            ///read cache
            pre=now;
            now=(now+1)%10;
            dy[now]=dy[pre];
            memcpy(day_sum[now],day_sum[pre],sizeof(day_sum[0]));
            ///7 day ago    ....
            while(scanf("%s",cache))
            {
                if(cache[0]=='<') break;
                if(strlen(cache)<4) continue;
                string word=cache;
                if(Hash[word]==0)
                {
                    rHash[id]=word;
                    Hash[word]=id++;
                }
                int ID=Hash[word];
                if(day_sum[pre][ID]==0)
                    dy[now].push_back(ID);
                day_sum[now][ID]++;
            }
        }
        else if(strcmp(cache,"