1121. Damn Single (25)-PAT甲级真题

“Damn Single (单身狗)” is the Chinese nickname for someone who is being single. You are supposed to find those who are alone in a big party, so they can be taken care of.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=50000), the total number of couples. Then N lines of the couples follow, each gives a couple of ID’s which are 5-digit numbers (i.e. from 00000 to 99999). After the list of couples, there is a positive integer M (<=10000) followed by M ID’s of the party guests. The numbers are separated by spaces. It is guaranteed that nobody is having bigamous marriage (重婚) or dangling with more than one companion.

Output Specification:

First print in a line the total number of lonely guests. Then in the next line, print their ID’s in increasing order. The numbers must be separated by exactly 1 space, and there must be no extra space at the end of the line.

Sample Input:

11111 22222
33333 44444
55555 66666
55555 44444 10000 88888 22222 11111 23333

Sample Output:

10000 23333 44444 55555 88888

分析: 设立数组couple[i] = j表示i的对象是j。一开始先设置为都是-1。设立数组isExist表示某人的对象是否来到了派对上。接收数据的时候,对于每一对a和b,将couple的a设置为b,b设置为a,表示他俩是一对。对于每一个需要判断的人,将其存储在guest数组里面,如果它不是单身的(也就是如果它的couple[guest[i]] != -1)那么就将它对象的isExist设置为1,表示他对象的对象(也就是他自己)来到了派对。这样所有isExist不为1的人,对象是没有来到派对的。把所有的人遍历后插入一个集合set里面,set的size就是所求的人数,set里面的所有数就是所求的人的递增排列~~~


❤ 点击这里 -> 订阅《PAT | 蓝桥 | LeetCode学习路径 & 刷题经验》by 柳婼

❤ 点击这里 -> 订阅《从放弃C语言到使用C++刷算法的简明教程》by 柳婼

❤ 点击这里 -> 订阅PAT甲级乙级、蓝桥杯、GPLT天梯赛、LeetCode题解离线版