stl-map的一道很好的题目

Damn Single (25)

“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 (<=50000), 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 (<=10000) 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

思路：pat甲级的一道题目，个人的第一次看到用map容器的用法，map的用法主要用于标记，这道题个人感觉很不错，运用map，很好说明map的一些常用的用法。用法在代码注释。

#include <cstdio>  #include <cstring>  #include <vector>  #include <map>  #include <algorithm>  #include <iostream>  using namespace std;  int main(){      int n,count=0;      string a,b;      map<string,string> c;      map<string,int> d;      vector<string> re;      cin>>n;      for(int i=1;i<=n;i++){          cin>>a>>b;          c.insert(pair<string,string>(a,b));//对a和b进行插入，说明a是key，b是value；        c.insert(pair<string,string>(b,a));  //想反的在进行一个插入，a与b此时连在一起的感觉。    }      cin>>n;      for(int i=1;i<=n;i++){          cin>>a;          d[a]=1;  //对a进行标记；    }      int flag=0;      map<string,int> ::iterator it;      for(it=d.begin();it!=d.end();it++){          if(d.count(c[it->first])==0){ //count函数说明value有没有出现，c【}it-》first 】的值是夫妻对方；主要要明白这里的it->first。            count++;              re.push_back(it->first); //map中的it->first是            自动排序的。               }      }      cout<<count<<endl;      for(int i=0;i<re.size();i++){          cout<<re.at(i)<<(i==re.size()-1?'\n':' ');      }      return 0;  }