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

题目大意：

给定每个人的家庭成员和其自己名下的房产，请你统计出每个家庭的人口数、人均房产面积及房产套数。首先在第一行输出家庭个数（所有有亲属关系的人都属于同一个家庭）。随后按下列格式输出每个家庭的信息：家庭成员的最小编号 家庭人口数 人均房产套数 人均房产面积。其中人均值要求保留小数点后3位。家庭信息首先按人均面积降序输出，若有并列，则按成员编号的升序输出

分析：

用并查集。分别用两个结构体数组，一个data用来接收数据，接收的时候顺便实现了并查集的操作union，另一个数组ans用来输出最后的答案，因为要计算家庭人数，所以用visit标记所有出现过的结点，对于每个结点的父结点，people++统计人数。标记flag == true，计算true的个数cnt就可以知道一共有多少个家庭。排序后输出前cnt个就是所求答案～～

原文链接：https://blog.csdn.net/liuchuo/article/details/52200109

题解

#include <bits/stdc++.h>

using namespace std;
struct Data
{
    int id,fid,mid,num,area;
    int cid[10];
}Data[1005];
struct node
{
    int id,people;
    double num,area;
    bool flag=false;
}ans[10000];
int father[10000],vis[10000];
int findFather(int x){
    while(x!=father[x]){
        x=father[x];
    }
    return x;
}
void Union(int a,int b){
    int faA=findFather(a);
    int faB=findFather(b);
    if(faA>faB)
        father[faA]=faB;
    else if(faA<faB)
        father[faB]=faA;
}
bool cmp(node a,node b){
    if(a.area!=b.area) return a.area>b.area;
    else return a.id<b.id;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,k;
    cin>>n;
    for(int i=0;i<10000;i++){
        father[i]=i;
    }
    for(int i=0;i<n;i++){
        cin>>Data[i].id>>Data[i].fid>>Data[i].mid>>k;
        vis[Data[i].id]=true;
        if(Data[i].fid!=-1){
            Union(Data[i].id,Data[i].fid);
            vis[Data[i].fid];
        }
        if(Data[i].mid!=-1){
            Union(Data[i].id,Data[i].mid);
            vis[Data[i].mid];
        }
        for(int j=0;j<k;j++){
            cin>>Data[i].cid[j];
            Union(Data[i].id,Data[i].cid[j]);
            vis[Data[i].cid[j]]=true;
        }
        cin>>Data[i].num>>Data[i].area;
    }
    for(int i=0;i<n;i++){
        int id=findFather(Data[i].id);
        ans[id].id=id;
        ans[id].num+=Data[i].num;
        ans[id].area+=Data[i].area;
        ans[id].flag=true;
    }
    int cnt=0;
    for(int i=0;i<10000;i++){
        if(vis[i]=true)
            ans[findFather(i)].people++;
        if(ans[i].flag==true)
            cnt++;
    }
    for(int i=0;i<10000;i++){
        if(ans[i].flag){
            ans[i].num=(double)(ans[i].num*1.0/ans[i].people);
            ans[i].area=(double)(ans[i].area*1.0/ans[i].people);
        }
    }
    sort(ans,ans+10000,cmp);
    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);
    }
    return 0;
}