1121. Damn Single (25)

题目：

“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

解答：

//A1121#include<cstdio>#include<vector>#include<algorithm>using namespace std;int main(){    int n,temp1,temp2;    int flag[100010]={0};    int couple[100010]={0};    vector<int> ans;    scanf("%d",&n);    for(int i=0;i<n;i++){        scanf("%d%d",&temp1,&temp2);        couple[temp1]=temp2;        couple[temp2]=temp1;    }    int m,temp3;    scanf("%d",&m);    for(int i=0;i<m;i++){        scanf("%d",&temp3);        if(flag[couple[temp3]]==-1){            flag[temp3]=0;            flag[couple[temp3]]=0;            continue;        }        flag[temp3]--;    }    for(int i=0;i<100010;i++){        if(flag[i]==-1) ans.push_back(i);    }    sort(ans.begin(),ans.end());    printf("%d\n",ans.size());    for(int i=0;i<ans.size();i++){        if(i!=0)printf(" ");        printf("%05d",ans[i]);//不按照输出格式写的话会有一个测试点出错     }    return 0;}

虽然通过了，但是觉可能是侥幸？

因为couple[100010]={0}。不知道疑虑是否确有其事