PAT甲级 1121. Damn Single (25)

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

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题目的意思是先给出n组数，每组的两个数表示一对在查询每个人，如果一个人没对象或者他对象不再查询里就把他输出，输出按编号从小到大输出

思路：开数组记录每个人的对象，把查询的人先标记一下，再把对象没标记或没对象的人扔进set里输出

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
int link[1000006];
bool flag[1000006];
int a[1000006];
int main()
{
    int n,x,y,m;
    scanf("%d",&n);
    memset(link,-1,sizeof link);
    for(int i=-0; i<n; i++)
    {
        scanf("%d%d",&x,&y);
        link[x]=y;
        link[y]=x;
    }
    scanf("%d",&m);
    memset(flag,0,sizeof flag);
    for(int i=0; i<m; i++)
    {
        scanf("%d",&a[i]);
        flag[a[i]]=1;
    }
    set<int>s;
    for(int i=0; i<m; i++)
    {
        if(flag[link[a[i]]]==0||link[a[i]]==-1)
            s.insert(a[i]);
    }

    printf("%d\n",s.size());
    if(s.size()==0)
        return 0;
    set<int>::iterator it=s.begin();
    int q=0;
    for(; it!=s.end(); it++)
    {
        if(q++)
            printf(" ");
        printf("%05d",*it);
    }
    printf("\n");
    return 0;
}