标签：2-2 ted 没有 ack tee sts each return item
"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.
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.
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.
3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333
5
10000 23333 44444 55555 88888
题意：找出谁是“单身狗”。。。先输入各对CP，然后再输入参加party的人员，统计输入的这些人里哪些是没有CP一起来的，升序输出。
分析：水题。。我思路是用一个数组couple，初始化为-1（因为人的编号为00000-99999），作用是来存储某人的CP，如11111的CP是22222，那么couple[11111]=22222,couple[22222]=11111。最后因为要格式输出，不知道迭代器怎么格式输出。。就先把结果存到了vector中，具体见代码 1 /**
2 * Copyright(c)
3 * All rights reserved.
4 * Author : Mered1th
5 * Date : 2019-02-27-21.51.47
6 * Description : A1121
7 */
8 #include<cstdio>
9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 const int maxn=100010;
20 int couple[maxn]={-1};
21 int main(){
22 #ifdef ONLINE_JUDGE
23 #else
24 freopen("1.txt", "r", stdin);
25 #endif
26 int n,m,u,v,t;
27 scanf("%d",&n);
28 for(int i=0;i<n;i++){
29 scanf("%d%d",&u,&v);
30 couple[u]=v;
31 couple[v]=u;
32 }
33 scanf("%d",&m);
34 vector<int> tem;
35 for(int i=0;i<m;i++){
36 scanf("%d",&t);
37 tem.push_back(t);
38 }
39 set<int> ans;
40 for(int i=0;i<m;i++){
41 int coup=couple[tem[i]];
42 if(coup==-1) ans.insert(tem[i]);
43 else{
44 if(find(tem.begin(),tem.end(),coup)==tem.end()){
45 ans.insert(tem[i]);
46 }
47 }
48 }
49 cout<<ans.size()<<endl;
50 vector<int> res;
51 for(auto it=ans.begin();it!=ans.end();it++){
52 res.push_back(*it);
53 }
54 for(int i=0;i<res.size();i++){
55 printf("%05d",res[i]);
56 if(i!=res.size()-1) printf(" ");
57 }
58 return 0;
59 }
标签：2-2 ted 没有 ack tee sts each return item
原文地址：https://www.cnblogs.com/Mered1th/p/10447948.html
