PAT-A 真题 – 1121 Damn Single

"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 ( $\leq$ 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 ( $\leq$ 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 (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:

3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333

Sample Output:

5
10000 23333 44444 55555 88888

题目大意：给定配偶列表，接着给出一堆人，找到没有配偶的人。注意在表中没出现的人就算没有配偶。把找到的单身狗排序输出。

这道题可以把读入的配偶表存放在一个表中，并建立双向映射的关系，然后对于给的人，一一在表中找是否有配偶，没有配偶就放到set中，有配偶就不要放进去，并且在set中找他的配偶，如果之前放进去过，就得删掉。

set自带去重排序功能，直接输出就好。

不会set的看这里：https://www.mmuaa.com/post/b9a637bded62f2a6.html

#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <cstdio>
using namespace std;

int G[100000];
set<int> S;

int main(){
  fill(G, G+100000, -1);
  int cnt;
  cin >> cnt;
  for(int i = 0; i < cnt; i++){
    int l, r;
    cin >> l >> r;
    G[l] = r, G[r] = l;
  }
  cin >> cnt;
  for(int i = 0; i < cnt; i++){
    int n;
    cin >> n;
    set<int>::iterator it = S.find(G[n]);
    if(G[n] != -1 && it != S.end()){
      S.erase(it);
    }
    else S.insert(n);
  }
  cout << S.size() << endl;
  for(set<int>::iterator it = S.begin(); it != S.end(); it++){
    if(it != S.begin()) cout << ' ';
    printf("%05d", *it);
  }
  return 0;
}

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