1121 Damn Single （25 分）

发布时间：2020-12-14 02:00:03 所属栏目：Linux 来源：网络整理

导读：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 conta

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 (

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 22222 22222 33333 44444 55555 66666 7 55555 44444 10000 88888 22222 22222 23333

Sample Output:

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22222的CP是22222，那么couple[22222]=22222,couple[22222]=22222。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 }

（编辑：李大同）

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