PAT A1121 Damn Single （25 分）——set遍历

"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
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333

Sample Output:

5
10000 23333 44444 55555 88888

 

#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
using namespace std;
const int maxn=100031;
int couple[maxn]={0};
vector<int> v;
int peop[maxn]={0};
int n;
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        int cp,cp2;
        scanf("%d %d",&cp,&cp2);
        couple[cp]=cp2;
        couple[cp2]=cp;
    }
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        int p;
        scanf("%d",&p);    
        peop[p]=1;
        v.push_back(p);
    }
    for(int i=0;i<n;i++){
        if(peop[v[i]]==1){
            if(peop[couple[v[i]]]==1 && couple[couple[v[i]]]==v[i]){
                peop[v[i]]=0;
                peop[couple[v[i]]]=0;
            }
        }
    }
    int cnt=0;
    for(int i=0;i<n;i++){
        if(peop[v[i]]==1){
            cnt++;
        }
    }
    printf("%d
",cnt);
    int num=0;
    sort(v.begin(),v.end());
    for(int i=0;i<n;i++){
        if(peop[v[i]]==1){
            printf("%05d",v[i]);
            num++;
            if(num!=cnt)printf(" ");
        }
    }
}

注意点：一开始想用set和map，发现map其实就是一个大数组，还不如直接开个大数组。一开始题目理解错了，以为只要找给的人里是单身狗的就行，结果是有对象的对象没来也算单身，行吧