B - Rebranding

一家小而骄傲的公司决定通过一系列营销策略进行品牌重塑,包括更改公司名称。公司聘请了多位设计师,每位设计师都会根据特定规则修改公司名称,最终形成新的品牌名称。本文介绍了这一过程,并提供了计算新名称的算法实现。

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B - Rebranding


The 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 x i by y i, and all the letters y i by x i. 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 x i coincides with y i. 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 x i and y i.

Output
Print the new name of the corporation.

Examples
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

使用char a[26]表示a-z 并存储将要替换成的字母,再根据这个序列对应修改s字符串

#include <iostream>
using namespace std;
char s[200000],a[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'},b,c,d;
int main()
{
	int n , m , i , j ,k , f1 , f2 ;
	cin >> n >> m ;
	getchar();
	gets (s);
	for (i = 0 ; i < m ; i++ )
	{
		cin >> b ;
		cin >> c ;
		for ( j = 0 ; j < 26 ; j++ )
		{
			if (a[j] == b) f1 = j ;
			if (a[j] == c) f2 = j ;
		}
		d = a [f1] ;
		a [f1] = a[f2] ;
		a [f2] = d ;
	}
	for(i = 0 ; i < n ; i++ )
	{
		s[i] = a[s[i] - 'a'] ;
	}
	cout << s ;
} 
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