codeforces 591B Rebranding

B. Rebranding

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard outputThe name of one small but proud corporation consists of n lowercase English letters. The Corporation has decided to try rebranding — an active marketing strategy, that includes a set of measures to change either the brand (both for the company and the goods it produces) or its components: the name, the logo, the slogan. They decided to start with the name.

For this purpose the corporation has consecutively hired m designers. Once a company hires the i-th designer, he immediately contributes to the creation of a new corporation name as follows: he takes the newest version of the name and replaces all the letters xi by yi, and all the letters yi byxi. This results in the new version. It is possible that some of these letters do no occur in the string. It may also happen that xi coincides with yi. The version of the name received after the work of the last designer becomes the new name of the corporation.

Manager Arkady has recently got a job in this company, but is already soaked in the spirit of teamwork and is very worried about the success of the rebranding. Naturally, he can't wait to find out what is the new name the Corporation will receive.

Satisfy Arkady's curiosity and tell him the final version of the name.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the length of the initial name and the number of designers hired, respectively.

The second line consists of n lowercase English letters and represents the original name of the corporation.

Next m lines contain the descriptions of the designers' actions: the i-th of them contains two space-separated lowercase English letters xi and yi.

Output

Print the new name of the corporation.

Sample test(s)

input

6 1
police
p m

output

molice

input

11 6
abacabadaba
a b
b c
a d
e g
f a
b b

output

cdcbcdcfcdc

//題意，輸入長度爲n的字符串，再輸入m組字符，若字符串中遇到每組的兩個字符則相互替換；此題直接暴力會超時，對輸入的m組數據處理，每輸入一組，同時處理一次；

思想是用棧保存下來一個字符最終會變成的字符；

#include "iostream"
#include "cstring"
#include "string"
#include "stack"
using namespace std;

int main(){
  string s;
  int n,m;
  char a,b;
  while(cin >> n >> m) {
       cin >> s;
       stack<int>r[26]; //製造一個棧，可以記錄26個字母
       for(int i = 0; i < 26; i++) r[i].push(i); //進行編號

        for(int i = 0; i < m; i++) {
            cin >> a >> b;
            for(int i = 0; i < 26; i++) {  
                if(r[i].top() == a - 'a')
                    r[i].push(b-'a');
                else if(r[i].top() == b - 'a')
                    r[i].push(a-'a');
            }
        }
      for(int i = 0; i < s.size(); i++) {
          cout << (char)(r[ s[i] - 'a' ].top() + 'a');
      }
      cout << endl;
}
}