PAT 1121 Damn Single

1121 Damn Single

"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

总结：这道题目已经在乙级写过了，但是第一次提交还是有点问题的，就是在最后打印的时候忘了id是从00000开始的，所以打印需要%05d打印00000这个id

代码：

#include <iostream>
#include <map>
#include <algorithm>
using namespace std;int a[100000]={-1};
bool st[100000];
int main(){int n,k,t,cot=0,index=0;map<int,int> m;scanf("%d",&n);for(int i=0;i<n;i++){int l,r;scanf("%d%d",&l,&r);a[l]=r;a[r]=l;}scanf("%d",&k);int b[k];for(int i=0;i<k;i++){scanf("%d",&b[i]);if(st[b[i]]){cot++;m[b[i]]=1;m[a[b[i]]]=1;}if(a[b[i]]!=-1)st[a[b[i]]]=true;}printf("%d\n",k-cot*2);sort(b,b+k);for(int i=0;i<k;i++){if(!m[b[i]]){if(index!=0)    printf(" ");printf("%05d",b[i]);index++;}}return 0;
}

好好学习，天天向上！

我要考研！