Codeforces 784B Santa Claus and Keyboard Check

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B. Santa Claus and Keyboard Check

Input file: standard input

Output file: standard output

Time limit: 2 second

Memory limit: 256 megabytes

 

Santa Claus decided to disassemble his keyboard to clean it. After he returned all the keys back, he suddenly realized that some pairs of keys took each other's place! That is, Santa suspects that each key is either on its place, or on the place of another key, which is located exactly where the first key should be.

In order to make sure that he's right and restore the correct order of keys, Santa typed his favorite patter looking only to his keyboard.

You are given the Santa's favorite patter and the string he actually typed. Determine which pairs of keys could be mixed. Each key must occur in pairs at most once.

Input

The input consists of only two strings s and t denoting the favorite Santa's patter and the resulting string. s and t are not empty and have the same length, which is at most 1000. Both strings consist only of lowercase English letters.

 

Output

If Santa is wrong, and there is no way to divide some of keys into pairs and swap keys in each pair so that the keyboard will be fixed, print «-1» (without quotes).

Otherwise, the first line of output should contain the only integer k (k ≥ 0) — the number of pairs of keys that should be swapped. The following k lines should contain two space-separated letters each, denoting the keys which should be swapped. All printed letters must be distinct.

If there are several possible answers, print any of them. You are free to choose the order of the pairs and the order of keys in a pair.

Each letter must occur at most once. Santa considers the keyboard to be fixed if he can print his favorite patter without mistakes.

 

Example

Input

helloworld
ehoolwlroz

Output

3
h e
l o
d z

 

Input

hastalavistababy
hastalavistababy

Output

0

 

Input

merrychristmas
christmasmerry

Output

-1

 

 

题目描述：

主人公拆了键盘的键帽进行清洁。当主人公把键帽装回键盘时，发现有些键帽装错了地方。为了恢复这些装错键帽的原来位置，主人公对着键盘打了一些字符。现在给出主人公对着键盘打的字符和实际屏幕显示的字符，判断哪些键帽的位置“调转”了。

 

题目分析：

这道题关键就是看清楚题：输出能通过交换两个键的键帽来恢复键盘正常键位的组合，如果不能通过交换就输出-1。理解了这个后，我们可以看看样例：第一个样例第一个字母主人公打了h，实际显示了e。如果要符合题意的话就必须除了打入h，显示e外，还要打入e，显示h，才能符合题目的输出要求。这个我们可以用c++ map来建立这种双向的映射关系。如果后面的组合不符合这种映射关系就输出-1，如果符合就输出之前记录的映射关系。我们可以用数组保存这些映射关系（只需要保存一个字母，就可以通过map来访问另一个字母）。

 

 

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <set>
#include <map>
#include <algorithm>
#include <utility>
#include <vector>
#include <queue>
using namespace std;

map<int, int> mymap;
int vec[30];   //保存所有的映射关系
int cnt = 0;   //计数

int main(){
    string s, t;
    cin >> s >> t;
    int n = s.length();

    for(int i = 0; i < n; i++){
            if(mymap[s[i]] == 0 && mymap[t[i]] == 0) {
                mymap[s[i]] = t[i];   //建立双向映射关系
                mymap[t[i]] = s[i];

                if(s[i] != t[i]){  //不相等就建立映射关系(相等就不用修复了,看题)
                    vec[++cnt] = s[i];
                }
            }
            else if(mymap[s[i]] != t[i] || mymap[t[i]] != s[i]){  //映射关系不对应
                cout << -1 << endl;
                return 0;
            }
    }

    cout << cnt << endl;
    for(int i = 1; i <= cnt; i++){
        printf("%c %c
", vec[i], mymap[vec[i]]);
    }
    return 0;
}