Source:

PAT A1121 Damn Single (25 分)

Description:

"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 (≤ 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 (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

Keys:

Code:

 /*
Data: 2019-08-14 16:32:45
Problem: PAT_A1121#Damn Single
AC: 13:30 题目大意:
找出独自前往派对的人
输入:
第一行给出,伴侣数N<=5e4(意味着最多1e5个人)
接下来N行,p1,p2
接下来一行,出席人数M<=1e4
接下来一行,给出M个id(5位)
输出:
独自参加派对的人数
id递增 基本思路:
有一点需要注意下,id值可能为0,因此哈希表的初始值置-1就可以了;
*/
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int M=1e5;
int cp[M],mp[M]={};
vector<int> pt,sg; int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE int n,v1,v2;
scanf("%d", &n);
fill(cp,cp+M,-);
for(int i=; i<n; i++)
{
scanf("%d%d", &v1,&v2);
cp[v1]=v2;
cp[v2]=v1;
}
scanf("%d", &n);
while(n--)
{
scanf("%d", &v1);
pt.push_back(v1);
mp[v1]=;
}
for(int i=; i<pt.size(); i++)
if(cp[pt[i]]==- || mp[cp[pt[i]]]==)
sg.push_back(pt[i]);
sort(sg.begin(),sg.end());
printf("%d\n", sg.size());
for(int i=; i<sg.size(); i++)
printf("%05d%c", sg[i],i==sg.size()-?'\n':' '); return ;
}

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