PTA 1121 Damn Single

题目链接：1121 Damn Single (25 分)

“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:

1 2	5 10000 23333 44444 55555 88888

题意

给定情侣关系，然后给出一次宴会上出席的人，问这些人中有多少单身狗 (伴侣没来的也算)。

思路

记录情侣关系，然后找出出席的人中不在情侣关系中的人以及伴侣没有来的人即可。

注意输出要补零，最后一行不用换行。

代码

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;

int couple[maxn] = {-1};

int main() {
    int n, m;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) {
        int x, y;
        scanf("%d%d", &x, &y);
        couple[x] = y;
        couple[y] = x;
    }
    scanf("%d", &m);
    vector<int> party;
    for(int i = 0; i < m; ++i) {
        int x;
        scanf("%d", &x);
        party.push_back(x);
    }
    set<int> ans;
    for(int i = 0; i < m; ++i) {
        int tmp = couple[party[i]];
        if(tmp == -1) {
            ans.insert(party[i]);
        } else {
            if(find(party.begin(), party.end(), tmp) == party.end()) {
                ans.insert(party[i]);
            }
        }
    }
    cout << ans.size() << endl;
    for (auto it = ans.begin(); it != ans.end(); ++it) {
        if(it != ans.begin()) {
            printf(" ");
        }
        printf("%05d", *it);
    }
    printf("\n");
    return 0;
}