1121. Damn Single (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

"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: 

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

Sample Output: 

5
10000 23333 44444 55555 88888

#include<cstdio>
#include<set>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 100000 + 10;
int couple[maxn];
int main()
{
  int N;
  fill(couple, couple + maxn, -1);
  scanf("%d", &N);
  int a, b;
  for (int i = 0; i < N; i++)
  {
    scanf("%d %d", &a, &b);
    couple[a] = b;
    couple[b] = a;
  }
  int M;
  scanf("%d", &M);
  set<int> single;
  int t;
  vector<int> v;
  for (int i = 0; i < M; i++)
  {
    scanf("%d", &t);
    v.push_back(t);
  }
  for (int i = 0; i < M; i++)
  {
    if (couple[v[i]] == -1)
    {
      single.insert(v[i]);
    }
    else
    {
      
      bool flag = false;
      for (int j = 0; j < M; j++)
      {
        if (v[j] == couple[v[i]])
        {
          flag = true;
          break;
        }
      }
      if (!flag)
      {
        single.insert(v[i]);
      }
    }
  }
  printf("%d\n", single.size());
  for (set<int>::iterator it = single.begin(); it != single.end(); it++)
  {
    if (it == single.begin())
      printf("%05d", *it);
    else
      printf(" %05d", *it);
  }
  return 0;
}