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77 lines
2.4 KiB
C
77 lines
2.4 KiB
C
/**
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* 1065. 单身狗(25)
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*
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* “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，
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* 以便给予特殊关爱。
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*
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* 输入格式：
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*
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* 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
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* 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
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* 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
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* 的ID，以空格分隔。题目保证无人重婚或脚踩两条船。
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*
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* 输出格式：
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*
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* 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
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* 空格分隔，行的首尾不得有多余空格。
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*
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* 输入样例：
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* 3
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* 11111 22222
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* 33333 44444
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* 55555 66666
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* 7
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* 55555 44444 10000 88888 22222 11111 23333
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* 输出样例：
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* 5
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* 10000 23333 44444 55555 88888
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*/
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#include <stdio.h>
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#define BLANK -1
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#define SIGNED -2
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#define SINGLE -3
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int main()
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{
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int couple[100000], count = 0, N, M, ID1, ID2;
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for(int i = 0; i < 100000; i++)
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couple[i] = BLANK;
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/* Read 'couple-list', every pair of 'index' and 'value' are a couple. */
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scanf("%d", &N);
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for(int i = 0; i < N; i++)
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{
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scanf("%d %d", &ID1, &ID2);
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couple[ID1] = ID2;
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couple[ID2] = ID1;
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}
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scanf("%d", &M);
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for(int i = 0; i < M; i++) /* Read guest list. */
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{
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scanf("%d", &ID1);
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if(couple[ID1] >= 0) /* If one has a mate */
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couple[ID1] = SIGNED; /* set SIGNED */
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else /* Else: not in the 'couple-list' */
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couple[ID1] = SINGLE, count++; /* set SINGLE */
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}
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/* If couple[ID] is >= 0 (but not signed) but couple[couple[ID]] == SIGNED
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* (signed in), this means 'ID' didn't come but his/her mate did.
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* So his/her mate is alone in the party, set to SINGLE. */
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for(int i = 0; i < 100000; i++)
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if(couple[i] >= 0 && couple[couple[i]] == SIGNED)
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couple[couple[i]] = SINGLE, count++;
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/* Those whose value is SINGLE is either a bachelor or came alone */
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printf("%d\n", count);
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for(int i = 0; i < 100000; i++)
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if(couple[i] == SINGLE)
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printf("%05d%c", i, --count ? ' ' : '\0');
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return 0;
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}
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