【笨方法學PAT】1121 Damn Single (25 分)
一、題目
"Damn Single (單身狗)" 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
二、題目大意
跟男女朋友名單和出席的人,判斷單身狗和只有一個人的人的名單。
三、考點
陣列
四、注意
1、使用陣列儲存情侶關係;
2、單身或者TA沒有參加的都列為單身狗。
五、程式碼
#include<iostream> #include<map> #include<set> #include<vector> #define N 100001 using namespace std; int n1[N] = { -1 }; bool show[N] = { false }; int main() { //read int n; cin >> n; for (int i = 0; i < n; ++i) { int a, b; cin >> a >> b; n1[a] = b; n1[b] = a; } int m; cin >> m; vector<int> v(m); set<int> sset; for (int i = 0; i < m; ++i) { cin >> v[i]; show[v[i]] = true; } //solve for (int i = 0; i < m; ++i) { //damn if (n1[v[i]] == -1) { sset.insert(v[i]); continue; } //just one if(show[n1[v[i]]]==false) { sset.insert(v[i]); continue; } } //output cout << sset.size() << endl; for (auto it = sset.begin(); it != sset.end(); ++it) { if (it != sset.begin()) cout << " "; printf("%05d", *it); } system("pause"); return 0; }