zoukankan      html  css  js  c++  java
  • Codeforces Round #389 Div.2 B. Santa Claus and Keyboard Check

    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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.

    Examples
    input
    helloworld
    ehoolwlroz
    output
    3
    h e
    l o
    d z
    input
    hastalavistababy
    hastalavistababy
    output
    0
    input
    merrychristmas
    christmasmerry
    output
    -1

    开个link数组,判断字符与字符间的对应关系。

    如果所有不同的字符对都满足双向对应,那么有解。

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<algorithm>
     4 #include<cstring>
     5 #include<queue>
     6 using namespace std;
     7 const int mxn=1010;
     8 int read(){
     9     int x=0,f=1;char ch=getchar();
    10     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    11     while(ch>='0' && ch<='9'){x=x*10-'0'+ch;ch=getchar();}
    12     return x*f;
    13 }
    14 char link[mxn];
    15 char s[mxn];
    16 char c[mxn];
    17 int k=0;
    18 char ans1[mxn],ans2[mxn];
    19 bool vis[mxn];
    20 int main(){
    21     int i,j;
    22     scanf("%s%s",s,c);
    23     int len=strlen(s);
    24     if(len!=strlen(c)){printf("-1
    ");return 0;}
    25     for(i=0;i<len;i++){
    26         if(c[i]!=link[s[i]]){
    27             if(!link[s[i]]){
    28                 link[s[i]]=c[i];
    29             }
    30             else{printf("-1
    ");return 0;}
    31         }
    32         if(s[i]!=link[c[i]]){
    33             if(!link[c[i]]){
    34                 link[c[i]]=s[i];
    35             }
    36             else {printf("-1
    ");return 0;}
    37         }
    38     }
    39     for(i=0;i<26;i++){
    40         int tmp='a'+i;
    41         if(!link[tmp])continue;
    42         if(link[tmp]!=tmp && !vis[tmp]){
    43             ans1[++k]=tmp;
    44             ans2[k]=link[tmp];
    45             vis[tmp]=vis[link[tmp]]=1;
    46         }
    47     }
    48     printf("%d
    ",k);
    49     for(i=1;i<=k;i++){
    50         printf("%c %c
    ",ans1[i],ans2[i]);
    51     }
    52     return 0;
    53 }
  • 相关阅读:
    关于PPTP不能打开部分网页
    在MarS Board上搭建PPTP
    Mars Board上无法使用apt-get
    在MarS board上烧录系统镜像
    PHP-变量(1)
    在KEIL 4.72中使用STM32的3.5固件库
    android SDK中java环境变量配置
    android SDK中打开AVD时提示PANIC: Could not open:XX
    ckplayer通过Mod-H264支持随意拖动功能
    430学习笔记-内置ADC12
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/6224282.html
Copyright © 2011-2022 走看看