【PAT甲级 - C++题解】1114 Family Property

✍个人博客：https://blog.csdn.net/Newin2020?spm=1011.2415.3001.5343
📚专栏地址：PAT题解集合
📝原题地址：题目详情 - 1114 Family Property (pintia.cn)
🔑中文翻译：家产
📣专栏定位：为想考甲级PAT的小伙伴整理常考算法题解，祝大家都能取得满分！
❤️如果有收获的话，欢迎点赞👍收藏📁，您的支持就是我创作的最大动力💪

1114 Family Property

This time, you are supposed to help us collect the data for family-owned property. Given each person’s family members, and the estate（房产）info under his/her own name, we need to know the size of each family, and the average area and number of sets of their real estate.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤1000). Then N lines follow, each gives the infomation of a person who owns estate in the format:

ID Father Mother k Child1⋯Childk Mestate Area
1

where ID is a unique 4-digit identification number for each person; Father and Mother are the ID’s of this person’s parents (if a parent has passed away, -1 will be given instead); k (0≤k≤5) is the number of children of this person; Child_i’s are the ID’s of his/her children; M_estate is the total number of sets of the real estate under his/her name; and Area is the total area of his/her estate.

Output Specification:

For each case, first print in a line the number of families (all the people that are related directly or indirectly are considered in the same family). Then output the family info in the format:

ID M AVG_sets AVG_area

where ID is the smallest ID in the family; M is the total number of family members; AVG_sets is the average number of sets of their real estate; and AVG_area is the average area. The average numbers must be accurate up to 3 decimal places. The families must be given in descending order of their average areas, and in ascending order of the ID’s if there is a tie.

Sample Input:

10
6666 5551 5552 1 7777 1 100
1234 5678 9012 1 0002 2 300
8888 -1 -1 0 1 1000
2468 0001 0004 1 2222 1 500
7777 6666 -1 0 2 300
3721 -1 -1 1 2333 2 150
9012 -1 -1 3 1236 1235 1234 1 100
1235 5678 9012 0 1 50
2222 1236 2468 2 6661 6662 1 300
2333 -1 3721 3 6661 6662 6663 1 100
1
2
3
4
5
6
7
8
9
10
11

Sample Output:

3
8888 1 1.000 1000.000
0001 15 0.600 100.000
5551 4 0.750 100.000
1
2
3
4

题意

给定每个人的家庭成员以及他/她自己名字下的不动产（地产）信息，我们需要知道每个家庭的成员数，以及人均不动产面积和人均房产套数。

输入拥有房产的人员的信息，格式如下：

ID Father Mother k Child_1 ... child_k M_estate Area
1

输出一个家庭的相关房产信息，格式如下：

ID M AVG_sets AVG_area
1

按人均房产面积降序顺序输出所有家庭信息。当存在人均房产面积相同的情况时，按 ID 升序顺序排序。

思路

具体思路如下：

1. 用一个结构体数组 e 来存储所有关系，并用一个结构体 family 来存储每个家庭的相关信息，然后输入拥有房产的人员信息。
2. 初始化并查集，然后开始从头往后遍历每个成员关系，并更新并查集，划分出每个人员所属的家庭集合。
3. 统计存在的集合个数即家庭个数，放入答案数组 res 中。
4. 对答案数组进性排序，按人均房产面积降序顺序输出所有家庭信息。当存在人均房产面积相同的情况时，按 ID 升序顺序排序。
5. 输出结果，注意输出房产总面积以及人均房产面积时保留 3 位小数。

代码

#include
using namespace std;

const int N = 10010;
int n, cnt;
int p[N], c[N], hc[N], ha[N];
bool st[N];

struct edge
{
    int a, b;
}e[N];

struct family
{
    int id, c, hc, ha;

    //重载比较运算
    bool operator < (const family& f)const
    {
        // ha / c 是否等于 f.ha / f.c
        if (ha * f.c != f.ha * c)  return ha * f.c > f.ha * c;
        return id < f.id;
    }
};

//并查集模板
int find(int x)
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main()
{
    cin >> n;

    //输入每个家庭信息
    cnt = 0;
    for (int i = 0; i < n; i++)
    {
        int id, father, mother, k;
        cin >> id >> father >> mother >> k;

        st[id] = true;
        //判断是否有父母
        if (father != -1)  e[cnt++] = { id,father };
        if (mother != -1)  e[cnt++] = { id,mother };
        //输入孩子信息
        while (k--)
        {
            int x;
            cin >> x;
            e[cnt++] = { id,x };
        }

        cin >> hc[id] >> ha[id];
    }

    for (int i = 0; i < N; i++)    p[i] = i, c[i] = 1; //初始化并查集
    for (int i = 0; i < cnt; i++)  //合并家庭
    {
        int a = e[i].a, b = e[i].b;
        st[a] = true, st[b] = true;
        int pa = find(a), pb = find(b);
        if (pa != pb)
        {
            if (pa > pb)   swap(pa, pb);
            c[pa] += c[pb];
            hc[pa] += hc[pb];
            ha[pa] += ha[pb];
            p[pb] = pa;
        }
    }

    //统计答案
    vector<family> res;
    for (int i = 0; i < N; i++)
        if (st[i] && p[i] == i)
            res.push_back({ i,c[i],hc[i],ha[i] });

    //对答案进行排序
    sort(res.begin(), res.end());

    cout << res.size() << endl;
    for (auto& f : res)
        printf("%04d %d %.3lf %.3lf\n", f.id, f.c, (double)f.hc / f.c, (double)f.ha / f.c);

    return 0;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90