Algorithm

B. Facetook Priority Wall
time limit per test
2 seconds
memory limit per test
256 megabytes
Problem Link- https://codeforces.com/problemset/problem/75/B
input
standard input
output
standard output

Facetook is a well known social network website, and it will launch a new feature called Facetook Priority Wall. This feature will sort all posts from your friends according to the priority factor (it will be described).

This priority factor will be affected by three types of actions:

  • 1. "X posted on Y's wall" (15 points),
  • 2. "X commented on Y's post" (10 points),
  • 3. "X likes Y's post" (5 points).

X and Y will be two distinct names. And each action will increase the priority factor between X and Y (and vice versa) by the above value of points (the priority factor between X and Y is the same as the priority factor between Y and X).

You will be given n actions with the above format (without the action number and the number of points), and you have to print all the distinct names in these actions sorted according to the priority factor with you.

Input

The first line contains your name. The second line contains an integer n, which is the number of actions (1 ≤ n ≤ 100). Then n lines follow, it is guaranteed that each one contains exactly 1 action in the format given above. There is exactly one space between each two words in a line, and there are no extra spaces. All the letters are lowercase. All names in the input will consist of at least 1 letter and at most 10 small Latin letters.

Output

Print m lines, where m is the number of distinct names in the input (excluding yourself). Each line should contain just 1 name. The names should be sorted according to the priority factor with you in the descending order (the highest priority factor should come first). If two or more names have the same priority factor, print them in the alphabetical (lexicographical) order.

Note, that you should output all the names that are present in the input data (excluding yourself), even if that person has a zero priority factor.

The lexicographical comparison is performed by the standard "<" operator in modern programming languages. The line a is lexicographically smaller than the line b, if either a is the prefix of b, or if exists such an i (1 ≤ i ≤ min(|a|, |b|)), that ai < bi, and for any j (1 ≤ j < i) aj = bj, where |a| and |b| stand for the lengths of strings a and b correspondently.

Examples
input
Copy
ahmed
3
ahmed posted on fatma's wall
fatma commented on ahmed's post
mona likes ahmed's post
output
Copy
fatma
mona
input
Copy
aba
1
likes likes posted's post
output
Copy
likes
posted



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>

struct str{std::string name; long pr;};
bool comp(str a, str b){
    if(a.pr > b.pr){return true;}
    else if(a.pr == b.pr && a.name < b.name){return true;}
    return false;
}

int main(){

    std::string myname; getline(std::cin, myname);
    std::string sn; getline(std::cin, sn);
    long n(0); for(long p = 0; p < sn.size(); p++){n = 10 * n + (sn[p] - '0');}

    std::map<std::string, long> pm;

    while(n--){
        std::string s; getline(std::cin, s);
        std::string from(""), to("");
        long score(0);
        long ind(0); while(s[ind] != ' '){from += s[ind]; ++ind;}
        ++ind;
        if(s[ind] == 'p'){ind += 10; score = 15;}
        else if(s[ind] == 'c'){ind += 13; score = 10;}
        else if(s[ind] == 'l'){ind += 6; score = 5;}
        while(s[ind] != '\''){to += s[ind]; ++ind;}
        if(from == myname){pm[to] += score;}
        else if(to == myname){pm[from] += score;}
        else{pm[to] += 0; pm[from] += 0;} //ensure that they are in the map
    }

    std::vector<str> v;
    for(std::map<std::string, long>::iterator it = pm.begin(); it != pm.end(); it++){str x; x.name = it->first; x.pr = it->second; v.push_back(x);}
    sort(v.begin(), v.end(), comp);
    for(long p = 0; p < v.size(); p++){std::cout << v[p].name << std::endl;}

    return 0;
}
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
ahmed
3
ahmed posted on fatma's wall
fatma commented on ahmed's post
mona likes ahmed's post

Output

x
–
+
cmd
fatma
mona
Advertisements

Demonstration

Codeforces Solution-B. Facetook Priority Wall-Solution in C, C++, Java, Python

Previous
Codeforces solution 1080-B-B. Margarite and the best present codeforces solution
Next
CodeChef solution DETSCORE - Determine the Score CodeChef solution C,C+