mirror of
https://github.com/xlucn/PAT.git
synced 2026-08-18 18:20:00 +08:00
70 lines
2.2 KiB
C
70 lines
2.2 KiB
C
/**
|
|
* 1065. 单身狗(25)
|
|
*
|
|
* “单身狗”是中文对于单身人士的一种爱称。本题请你从上万人的大型派对中找出落单的客人，
|
|
* 以便给予特殊关爱。
|
|
*
|
|
* 输入格式：
|
|
*
|
|
* 输入第一行给出一个正整数N（<=50000），是已知夫妻/伴侣的对数；随后N行，每行给出一
|
|
* 对夫妻/伴侣——为方便起见，每人对应一个ID号，为5位数字（从00000到99999），ID间以空
|
|
* 格分隔；之后给出一个正整数M（<=10000），为参加派对的总人数；随后一行给出这M位客人
|
|
* 的ID，以空格分隔。题目保证无人重婚或脚踩两条船。
|
|
*
|
|
* 输出格式：
|
|
*
|
|
* 首先第一行输出落单客人的总人数；随后第二行按ID递增顺序列出落单的客人。ID间用1个
|
|
* 空格分隔，行的首尾不得有多余空格。
|
|
*
|
|
* 输入样例：
|
|
* 3
|
|
* 11111 22222
|
|
* 33333 44444
|
|
* 55555 66666
|
|
* 7
|
|
* 55555 44444 10000 88888 22222 11111 23333
|
|
* 输出样例：
|
|
* 5
|
|
* 10000 23333 44444 55555 88888
|
|
*/
|
|
#include <stdio.h>
|
|
int main()
|
|
{
|
|
int couple[100001] = {0}, N, ID1, ID2, M, count = 0;
|
|
|
|
/* every pair of 'index' and 'value' are a couple.
|
|
* record ID + 1 to avoid '00000' conflict with 0 */
|
|
scanf("%d", &N);
|
|
for(int i = 0; i < N; i++)
|
|
{
|
|
scanf("%d %d", &ID1, &ID2); ID1++; ID2++;
|
|
couple[ID1] = ID2;
|
|
couple[ID2] = ID1;
|
|
}
|
|
|
|
/* Record those who come. If one has a mate then set 0 (means signed in),
|
|
* else (means not even in the 'couple-list') set -1 (means bachelor) */
|
|
scanf("%d", &M);
|
|
for(int i = 0; i < M; i++)
|
|
{
|
|
scanf("%d", &ID1); ID1++;
|
|
if(couple[ID1]) couple[ID1] = 0;
|
|
else {couple[ID1] = -1; count++;}
|
|
}
|
|
|
|
/* If couple[ID] is positive but couple[couple[ID]] is zero,
|
|
* it means 'ID' didn't come but his/her mate did(signed in). */
|
|
for(int i = 0; i < 100001; i++) if(couple[i] > 0 && !couple[couple[i]])
|
|
{
|
|
couple[couple[i]] = -1;
|
|
count++;
|
|
}
|
|
|
|
/* Those whose value is -1 is a bachelor or came alone */
|
|
printf("%d\n", count);
|
|
for(int i = 0; i < 100001; i++) if(couple[i] == -1)
|
|
printf("%05d%c", i - 1, --count ? ' ' : '\0');
|
|
|
|
return 0;
|
|
}
|