机试练习（一）——Codeforces 784B Santa Claus and Keyboard Check

最近在准备机试，对练习的机试题做个总结。之前没有学过C++，只学过C语言，但是实际用起来的时候发现C++是更适合机试的语言，因为它的库函数更多，能支持更多操作，将一些代码简化。

习惯了C语言定义字符串char s[1000]; 在C++中一个string s就OK了等等。

题目

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

---

读题读了10分钟才完全懂，真是太慢了。。。

它的大意就是给你两个字符串，可能键帽按错了，导致某两个字母都颠倒了，让你给他恢复成原来的样子，如果能恢复，就输出错误个数，并输出所有错误的键对，如果恢复不了，就输出-1，如果一开始就是完全正确的，就输出0（这其实是能恢复的一种特殊情况，其实可以和第一个合并）。

代码

下面是我自己的代码，会在第14个样例出错，我太菜了，暂时还没有找到问题（加黑标注一下，如果网友发现我的代码哪里有问题，可以直接评论告诉我）

点击查看代码

#include<bits/stdc++.h>
using namespace std;

int main(){
	char s[1000],t[1000],o[2000],temp1,temp2;
	int flag[1000]={0};
	int sum=0;
	gets(s);
	gets(t);
	int slen = strlen(s);
	int i=0,j=0,k=0;
	if(strcmp(s,t)==0){
		cout<<0<<endl;
		return 0;
	}
	else{
		while(i<slen){
			if(s[i]==t[i]){
				++i;
			}
			else{
				temp1=s[i];
				temp2=t[i];
				for(j=i+1;j<slen;j++){
					if(t[j]==temp2&&s[j]!=temp1||s[j]==temp1&&t[j]!=temp2||t[j]==temp1&&s[j]!=temp2||s[j]==temp2&&t[j]!=temp1){
						cout<<-1<<endl;
						return 0;
					}
				}
				o[k++]=temp1;
				o[k++]=temp2;
				sum+=1;
				for(j=i;j<slen;j++){
					if(t[j]==temp2) flag[j]=1;
					if(t[j]==temp1){
						t[j]=temp2;
					}
				}
				for(j=i;j<slen;j++){
					if(flag[j]==1){
						t[j]=temp1;
						flag[j]=0;
					}
				}
			}
		}
		if(strcmp(s,t)==0){
			cout<<sum<<endl;
			for(i=0;i<k;i=i+2){
				cout<<o[i]<<" "<<o[i+1]<<endl;
			}
			return 0;
		}
	}
}

下面一个大佬的AC代码（来自：https://www.cnblogs.com/happy-MEdge/p/10455803.html ），看了之后非常敬佩，是我不会用map,其实我一直在想，怎么给他做映射，但是不会写，我可能更应该系统学一下C++，

思路

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）。

代码

点击查看代码

#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\n", vec[i], mymap[vec[i]]);
    }
    return 0;
}