stl-map的一道很好的题目
Damn Single (25)
“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 (<=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甲級(jí)的一道題目,個(gè)人的第一次看到用map容器的用法,map的用法主要用于標(biāo)記,這道題個(gè)人感覺很不錯(cuò),運(yùn)用map,很好說(shuō)明map的一些常用的用法。用法在代碼注釋。
#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));//對(duì)a和b進(jìn)行插入,說(shuō)明a是key,b是value;c.insert(pair<string,string>(b,a)); //想反的在進(jìn)行一個(gè)插入,a與b此時(shí)連在一起的感覺。} cin>>n; for(int i=1;i<=n;i++){ cin>>a; d[a]=1; //對(duì)a進(jìn)行標(biāo)記;} int flag=0; map<string,int> ::iterator it; for(it=d.begin();it!=d.end();it++){ if(d.count(c[it->first])==0){ //count函數(shù)說(shuō)明value有沒有出現(xiàn),c【}it-》first 】的值是夫妻對(duì)方;主要要明白這里的it->first。count++; re.push_back(it->first); //map中的it->first是自動(dòng)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; }總結(jié)
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