Pat 1121 damn single [simple]

1121 damn single (25 points)

"Damn single (single dog)" 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.

Input specification:

Each input file contains one test case. for each case, the first line gives a positive integer N (≤ 50,000), 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 (≤ 10,000) followed by m id's of the party guests. the numbers are separated by spaces. it is guaranteed that nobody is having bigamous marriage (remarriage) 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:

311111 2222233333 4444455555 66666755555 44444 10000 88888 22222 11111 23333

Sample output:

510000 23333 44444 55555 88888

Question: give n couples, and then give m to those who participate in the banquet to determine how many of them are not partners to attend the banquet (they are already married, but those without a partner will also be included)

# Include <iostream> # include <vector> # include <map> using namespace STD; Map <string, string> m2f; Map <string, string> indium; int main () {int N; CIN> N; string X, Y; For (INT I = 0; I <n; I ++) {CIN> x> Y; m2f [x] = y; m2f [y] = x;} int m; CIN> m; string s; For (INT I = 0; I <m; I ++) {CIN> S; indium [s] = 1; // because the map here is naturally sorted, the final vector is also naturally sorted. } Int Ct = 0; vector <string> VT; For (Auto it = indium. Begin (); it! = Indium. end (); It ++) {string STR = it-> first; If (Indium. count (m2f [STR]) = 0) {vt. push_back (STR) ;}} cout <vt. size () <'\ n'; For (INT I = 0; I <vt. size (); I ++) {cout <VT [I]; if (I! = Vt. Size ()-1) cout <";}return 0 ;}

 

// It is relatively simple, but I submitted it twice. Why?

1. The main output format must be 1! = Vt. Size ()-1, do you know?

Pat 1121 damn single [simple]