1121 Damn Single (25point(s)) - C語言 PAT 甲級

1121 Damn Single (25point(s))

“Damn Single (單身狗)” 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 (≤ 50,000), 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 (≤ 10,000) 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

題目大意：

1065 單身狗 (25point(s))

設計思路：

1065 單身狗（C語言）

1. 用 -1 當初始值，均爲單身狗（因爲有 0 號客人，所以最好不用 0 當初始值）
2. 用數組映射一對伴侶
3. 用 -2 表示客人到場。讀取客人入場，用映射尋找自己的伴侶，當自己單身或伴侶還未到場，自己用 -2 表示到場
4. 若映射的伴侶狀態爲 -2 到場，則兩人配對成功，兩人用 -3 表示脫離單身狗標籤，單身狗數目減二

編譯器：C (gcc)

#include <stdio.h>

#define BLANK -1
#define ATTEND -2
#define PAIR -3

int main()
{
        int n, m, couple[100000];
        int id1, id2, count = 0;
        int i;
        for (i = 0; i < 100000; i++)
                couple[i] = BLANK;

        scanf("%d", &n);
        for (i = 0; i < n; i++) {
                scanf("%d %d", &id1, &id2);
                couple[id1] = id2;
                couple[id2] = id1;
        }

        scanf("%d", &m);
        for (i = 0; i < m; i++) {
                scanf("%d", &id1);
                if (couple[id1] >= 0 && couple[couple[id1]] == ATTEND) {
                        couple[couple[id1]] = PAIR;
                        couple[id1] = PAIR;
                        count += 2;
                } else {
                        couple[id1] = ATTEND;
                }
        }

        printf("%d\n", m - count);
                count = m - count;
        for (i = 0; i < 100000; i++)
                if (couple[i] == ATTEND)
                        printf("%05d%s", i, --count? " " : "");

        return 0;
}

此方案二（同樣 AC）整體思路是，用正數記錄伴侶，伴侶到場後更新映射值爲相應的負數，這樣最後彼此的映射值還在。但有 0 號客人，他正負都是 0，要對此特別處理，故採取了較好處理的方案一（用 -1、-2、-3 表示不同的狀態）

#include <stdio.h>

int comp(const void *a, const void *b)
{
        return *(int*)a - *(int*)b;
}

int main()
{
        int n, m, couple[100001] = {0};
        int id1, id2, count = 0, id[100001];
        int i;

        scanf("%d", &n);
        for (i = 0; i < n; i++){
                scanf("%d %d", &id1, &id2);
                id1++;
                id2++;
                couple[id1] = id2;
                couple[id2] = id1;
        }

        scanf("%d", &m);
        for (i = 0; i < m; i++){
                scanf("%d", &id[i]);
                id[i]++;
                id2 = abs(couple[id[i]]);
                couple[id2] *= -1;
        }

        qsort(id, m, sizeof(id[0]), comp);

        for (i = 0; i < m; i++)
                if (couple[id[i]] >= 0)
                        count++;
        printf("%d\n", count);
        for (i = 0; i < m; i++)
                if (couple[id[i]] >= 0)
                        printf("%05d%s", id[i] - 1, --count ? " " : "");

        return 0;
}