1121. Damn Single (25)

曾经终败给现在 2022-05-28 12:37 145阅读 0赞

"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 (<=50000), 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 (<=10000) 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

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题目大意：

代码：

#include<stdio.h>
    #include<map>
    #include<vector>
    #include<algorithm>
    using namespace std;
    map<int,int> Map1,Map2;
    vector<int> v;
    int main()
    {
        int i,j,n,m,k,t;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&k,&t);
            Map1[k]=t;
            Map1[t]=k;
        }
        scanf("%d",&m);
        for(i=0;i<m;i++)
        {
            scanf("%d",&k);
            Map2[k]=1;
        }
        map<int,int>::iterator it;
        for(it=Map2.begin();it!=Map2.end();it++)
        {
            if(Map1.find(it->first)!=Map1.end())
            {
                if(Map2.find(Map1[it->first])==Map2.end())
                {
                  v.push_back(it->first);
                }
            }
            else
            {
               v.push_back(it->first);
            }
        }
        sort(v.begin(),v.end());
        printf("%d\n",v.size());
        for(i=0;i<v.size();i++)
        {
            if(i==0)
            {
                printf("%05d",v[i]);
            }
            else
            {
                printf(" %05d",v[i]);
            }
        }
        return 0;
    }