PAT（甲级）1121.Damn Single（25）
PAT 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.
输入格式：
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.
输出格式：
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.
输入样例：
3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333输出样例：
5
10000 23333 44444 55555 88888题目分析：乙级1065的一个题目，具体分析请看乙级1065
注意：输出格式为%05d，否则测试点3会出错。
AC代码：
#include <iostream>
#include <unordered_map>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 100000;
bool is_appear[maxn] = {false};
vector<int>v;
unordered_map<int, int> mp;
int main(){
//freopen("in.txt", "r", stdin);
int n, a, b;
scanf("%d", &n);
for(int i=0; i<n; ++i){
scanf("%d%d", &a, &b);
mp.insert(make_pair(a, b));
}
int m, tmp;
scanf("%d", &m);
for(int i=0; i<m; ++i){
scanf("%d", &tmp);
is_appear[tmp] = true;
}
for(auto it=mp.begin(); it!=mp.end(); ++it){
if(is_appear[it->first] && is_appear[it->second]){
is_appear[it->first] = is_appear[it->second] = false;
}
}
for(int i=0; i<maxn; ++i){
if(is_appear[i])
v.push_back(i);
}
sort(v.begin(), v.end());
for(int i=0; i<v.size(); ++i){
printf("%05d", v[i]);
if(i!=v.size()-1)printf(" ");
}
return 0;
}腾讯云开发者
