# 1114. Family Property (25)-PAT甲级真题（并查集）

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 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; Childi’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

#### 题目理解

• 给出每个家庭的信息和房产信息
• 本人ID
• 父亲ID
• 母亲ID
• 小孩个数
• 小孩id
• 拥有房产数
• 拥有房产面积
• 计算整个家族的平均房产信息
• 家庭最小成员编号
• 家庭成员个数
• 总房产数
• 平均房产数
• 平均面积
• 家庭信息按照平均面积进行降序排序，平均面积相同时按照序号升序输出，平均值保留3位小数

#### 分析

• 设置一个DATA结构体存储输入信息，设置ans结构体存储输出结果
• 设置UNION函数将同一个家族的成员进行合并，find函数用于查找root节点
• 设置比较规则cmp1用于结果排序，平均面积进行降序排序，平均面积相同时按照序号升序输出
• visit[10000]数组用于记录哪些id出现过

• 接收输入，并将id与mid，fid，cid[]进行union操作
• 统计各家族的房产数量和房产面积
• 统计各家族的人数（累加所有visit[i]==1，ans[find(i)].people++），统计家族个数（ans[i].flag==1, cnt++)
• 统计平均房产数和面积
• 将结果进行排序，使用cmp1规则
• 结果输出，平均数保留3位小数
#include
#include

using namespace std;
struct DATA {
int id, fid, mid, num, area;
int cid[10];
} data[1001];
struct node {
int id, people;
double num, area;
bool flag = false;  //root节点flag=true
} ans[10000];

int father[10000];
int visit[10000];

int find(int x) {
while (x != father[x]) {
x = father[x];
}
return x;
}

void Union(int a, int b) {
int faA = find(a);
int faB = find(b);
if (faA > faB)
father[faA] = faB;
else
father[faB] = faA;
}

//多规则排序
int cmp1(node a, node b) {
if (a.area != b.area)
return a.area > b.area;
else
return a.id < b.id;
}

int main() {
int n, k, cnt = 0;
cin >> n;
for (int i = 0; i < 10000; i++) {
father[i] = i;
}
//接收输入，并进行Union操作
for (int i = 0; i < n; i++) {
cin >> data[i].id >> data[i].fid >> data[i].mid >> k;
visit[data[i].id] = true;
if (data[i].fid != -1) {
visit[data[i].fid] = true;
Union(data[i].id, data[i].fid);
}
if (data[i].mid != -1) {
visit[data[i].mid] = true;
Union(data[i].id, data[i].mid);
}
for (int j = 0; j < k; j++) {
cin >> data[i].cid[j];
visit[data[i].cid[j]] = true;
Union(data[i].id, data[i].cid[j]);
}
cin >> data[i].num >> data[i].area;
}
//累计各家族的房产数量，面积
for (int i = 0; i < n; i++) {
int root = find(data[i].id);
ans[root].id = root;
ans[root].num += data[i].num; //累加家族房产数量
ans[root].area += data[i].area;  //累加家族面积
ans[root].flag = true;

}

for (int i = 0; i < 10000; i++) {
if (visit[i])
ans[find(i)].people++; //各家族人数统计
if (ans[i].flag)
cnt++;  //家族个数统计
}
//统计平均房产数和面积
for (int i = 0; i < 10000; i++) {
if (ans[i].flag) {
ans[i].num = (double) (ans[i].num / ans[i].people);
ans[i].area = (double) (ans[i].area / ans[i].people);
}
}
sort(ans, ans + 10000, cmp1);
cout << cnt << endl;
for (int i = 0; i < cnt; i++) {
printf("%04d %d %.3f %.3f\n", ans[i].id, ans[i].people, ans[i].num, ans[i].area);
}

}