题目

https://pintia.cn/problem-sets/994805342720868352/problems/994805352359378944

题意

给出n组朋友序列，再给出参加派对的m个人，升序输出m个人中没有朋友的人

代码解析

本题的关键就是使用set的find函数，如果直接遍历数组会超时

AC代码

#include<bits/stdc++.h>
using namespace std;
unordered_map<int,int> f;
set<int> s,ans;
int main()
{int n,m,x,y,t;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d %d",&x,&y);f[x]=y;f[y]=x;}scanf("%d",&m);for(int i=0;i<m;i++){scanf("%d",&t);s.insert(t);} for(auto it=s.begin();it!=s.end();it++){auto res=s.find(f[*it]);if(res==s.end()) ans.insert(*it); }cout<<ans.size()<<endl;for(auto it=ans.begin();it!=ans.end();it++){if(it!=ans.begin()) cout<<" ";printf("%05d",*it);}
}

参考

1121 Damn Single（25 分）

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PAT——1121 Damn Single 甲级

1121 Damn Single

题目

题意

代码解析

AC代码

参考

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