题意:
L<=106的字符串s,104种变换(a,b),每次变换所有的b字符到a字符,只变大写字母,求变换后的字符串
分析:
暴力变换26种字母,复杂度O(26R+L)
代码:
//
// Created by TaoSama on 2015-12-08
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
char s[N], cov[26];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
while(t--) {
scanf("%s", s);
for(int i = 0; i < 26; ++i) cov[i] = i + 'A';
int n; scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
char a, b; scanf(" %c %c", &a, &b);
for(int j = 0; j < 26; ++j)
if(cov[j] == b) cov[j] = a;
}
for(int i = 0; s[i]; ++i) {
if(isupper(s[i]))
s[i] = cov[s[i] - 'A'];
putchar(s[i]);
}
puts("");
}
return 0;
}