1114 Family Property

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 Child
​1
​​ ⋯Child
​k
​​ M
​estate
​​ Area

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
Sample Output:
3
8888 1 1.000 1000.000
0001 15 0.600 100.000
5551 4 0.750 100.000
分析：考察并查集的知识。
可以分成两步处理，第一步输入n条数据，记录id,fin,mid,child的根结点。
第二步统计家族个数，以及相应的人数。
参考代码（来源别处）

#include<iostream>
#include<vector>
#include<algorithm>
#define N 10002
using namespace std;
struct node {
    int id, fid, mid, estate;
    double area;
}person[N];
struct node1 {
    int id, people= 0, total_estate=0;
    double total_area = 0, AVG_area;
    bool flag = false;
}family[N];
bool visited[N];
int father[N];
int find(int x) {             //找根
    while (x != father[x])
        x = father[x];
    return x;
}
void Union(int a, int b) {   //并集！！！
    int findA = find(a);
    int findB = find(b);
    if (findA < findB)
        father[findB] = findA;
    else if(findB<findA)
        father[findA] = findB;
}
bool cmp(node1 a, node1 b) {
    if (a.AVG_area != b.AVG_area)
        return a.AVG_area > b.AVG_area;
    else return a.id < b.id;
}
int main()
{
    int n,child, k;
    for (int i = 0; i < N; i++) {
        father[i] = i;
        visited[i] = false;
    }
    scanf_s("%d", &n);

    for (int i = 0; i < n; i++) {
        scanf_s("%d%d%d%d", &person[i].id, &person[i].fid, &person[i].mid, &k);
        visited[person[i].id] = true;
        if (person[i].fid != -1) {
            visited[person[i].fid] = true;
            Union(person[i].id, person[i].fid);
        }
        if (person[i].mid != -1) {
            visited[person[i].mid] = true;
            Union(person[i].id, person[i].mid);
        }
        for (int j = 0; j < k; j++) {
            scanf_s("%d", &child);
            visited[child] = true;
            Union(person[i].id, child);
        }
        scanf_s("%d%lf", &person[i].estate, &person[i].area);
    }

    //n条记录
    for (int i = 0; i < n; i++) {
        int x = find(person[i].id);
        family[x].flag = true;
        family[x].id = x;
        family[x].total_area += person[i].area;
        family[x].total_estate += person[i].estate;
    }
    int cnt = 0;
    for (int i = 0; i < N; i++) {
        if (visited[i]) {
            int x = find(i);
            family[x].people++;
        }
        if (family[i].flag)
            cnt++;
    }
    for (int i = 0; i < N; i++) {
        if (family[i].people)
            family[i].AVG_area = family[i].total_area / family[i].people;
        else family[i].AVG_area = 0;
    }
    printf("%d\n", cnt);
    sort(family, family + N, cmp);
    for (int i = 0; i < cnt; i++) {
        printf("%04d %d %.3f %.3f\n", family[i].id, family[i].people, family[i].total_estate*1.0 / family[i].people, family[i].AVG_area);
    }


    return 0;
}

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