CF 591B Rebranding

最新推荐文章于 2022-10-23 14:46:37 发布

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最新推荐文章于 2022-10-23 14:46:37 发布

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分类专栏：模拟技巧题 (●´ω｀●)φ ACM水题训练

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本文探讨了一家小型企业通过连续聘请设计师进行字母替换，实现品牌重塑的过程。通过输入公司原始名称长度和设计师数量，以及设计师的具体操作（即字母替换），读者可以了解最终的企业名称演变。案例中包括了输入参数、操作流程和输出结果的详细说明，旨在揭示企业名称在品牌重塑过程中的变化路径。

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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 xi by yi, and all the letters yi by xi. 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 xicoincides 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

Note

就是字符替换题，这题直接用数组表示字母，模拟字母之间的交换，最后将结果全部付给字符串。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int num[40];
void init()
{
	for(int i=0;i<=25;i++)
	num[i] = i;
}
int main()
{
	int x,y;
	cin >> x >> y;
	string s;
	cin >> s;
	init();
	while(y--)
	{
		char a,b,aa,aaa;
		scanf("\n%c %c",&a,&b);
		//cout<<a<<b<<endl;
		for(int i=0;i<=25;i++)
		{
		int flag = 0;
		
		for(int j=0;j<=25;j++)
		{
			if(num[i]==a-'a'&&num[j]==b-'a')
			{
				flag = 1;
				swap(num[i],num[j]);
			}	
		}
		if(flag == 1)
		break;
	}
	}
	for(int i=0;i<x;i++)
	{
		s[i] = num[s[i]-'a']+'a';
	}
	cout<<s<<endl;
	return 0;
}