PAT (Advanced Level) Practise 1121 Damn Single (25)

1121. Damn Single (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

"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

题意：告诉你有n对夫妻，问在一个party上有几个单身狗（伴侣不在也算），并输出

解题思路：直接标记一下就好

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n,m;
int ans[100060];
int a[100060],visit[1000060],b[1000060];

int main()
{
while (~scanf("%d", &n))
{
memset(visit, 0, sizeof visit);
memset(b, -1, sizeof b);
int x, y;
for (int i = 1; i <= n; i++)
{
scanf("%d%d", &x, &y);
b[x] = y, b[y] = x;
}
int res = 0;
scanf("%d", &m);
for (int i = 1; i <= m; i++)
{
scanf("%d", &a[i]);
visit[a[i]] = 1;
}
for (int i = 1; i <= m; i++)
if (!visit[b[a[i]]] || b[a[i]] == -1) ans[res++] = a[i];
sort(ans, ans + res);
printf("%d\n", res);
if (res == 0) continue;
printf("%05d", ans[0]);
for (int i = 1; i < res; i++) printf(" %05d", ans[i]);
printf("\n");
}
return 0;
}