CodeForces 591B 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 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 xiand yi.

Output

Print the new name of the corporation.

Sample test(s)

input

6 1policep m

output

molice

input

11 6abacabadabaa bb ca de gf ab b

output

cdcbcdcfcdc

题意：给出一个字符串，现在有m次变换，每次变换需要两个字母a和b互换，问m次变换后新的字符串是什么。

因为只有26个字母，那么每次变换时将它们的位置变换一下，然后对应输出就行啦。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e6+10;
const int INF=0x3f3f3f3f;

char s[N];

int main ()
{
    int n, m, i;
    char a[5], b[5];
    int c[30]; ///c[i]表示‘a‘+i这个字母应该要变换成‘a‘+c[i]

    while (scanf("%d%d", &n, &m) != EOF)
    {
        scanf("%s", s);

        for (i = 0; i < 26; i++)
            c[i] = i;

        while (m--)
        {
            int aa, bb;

            scanf("%s %s", a, b);

            for (i = 0; i < 26; i++)
            {
                if (c[i] == a[0]-‘a‘)
                    aa = i;
                if (c[i] == b[0]-‘a‘)
                    bb = i;
            }

            swap(c[aa], c[bb]); ///找到与之对应的位置，互换
        }

        for (i = 0; s[i] != ‘\0‘; i++)
            printf("%c", c[s[i]-‘a‘] + ‘a‘);

        printf("\n");
    }

    return 0;
}