PAT 1121 Damn Single[简单]
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 (≤?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 22222 22222 33333 44444 55555 66666 7 55555 44444 10000 88888 22222 22222 23333 Sample Output:5 10000 23333 44444 55555 88888 ?题目大意:给出n对情侣,然后再给出m对参加宴会的人,判断这些人当中有多少人是没有伙伴来参加宴会的(本身已经结婚,但是没带伴侣来的也会被计算进来) #include <iostream> #include<vector> #include<map> using namespace std; map<string,string> m2f; map<string,string> inp; int main() { int n; cin>>n; string x,y; for(int i=0;i<n;i++){ cin>>x>>y; m2f[x]=y; m2f[y]=x; } int m; cin>>m; string s; for(int i=0;i<m;i++){ cin>>s; inp[s]=1;//因为这里的map是自然排序的,所以最终vector里也是自然排序的。 } int ct=0; vector<string> vt; for(auto it=inp.begin();it!=inp.end();it++){ string str=it->first; if(inp.count(m2f[str])==0){ vt.push_back(str); } } cout<<vt.size()<<‘n‘; for(int i=0;i<vt.size();i++){ cout<<vt[i]; if(i!=vt.size()-1)cout<<" "; } return 0; } ? //还是比较简单的,但是还是提交了两次,为什么呢? 1.需要主要最后的输出格式,是1!=vt.size()-1,知道了吗?? (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |