1121 Damn Single (25 point(s))
"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
经验总结:
emmmm 简单的划水题,考察的是hash的应用~
AC代码
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn=100010;
int couple[maxn]={-1},n;
bool isattend[maxn]={false};
int main()
{
int a,b;
scanf("%d",&n);
for(int i=0;i<n;++i)
{
scanf("%d%d",&a,&b);
couple[a]=b;
couple[b]=a;
}
vector<int> search,ans;
scanf("%d",&n);
search.resize(n);
for(int i=0;i<n;++i)
{
scanf("%d",&search[i]);
if(couple[search[i]]!=-1)
isattend[couple[search[i]]]=true;
}
for(int i=0;i<n;++i)
{
if(isattend[search[i]]==false)
ans.push_back(search[i]);
}
sort(ans.begin(),ans.end());
printf("%d\n",ans.size());
for(int i=0;i<ans.size();++i)
printf("%05d%c",ans[i],i<ans.size()-1?' ':'\n');
return 0;
}