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 (≤). 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) 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:

106666 5551 5552 1 7777 1 1001234 5678 9012 1 0002 2 3008888 -1 -1 0 1 10002468 0001 0004 1 2222 1 5007777 6666 -1 0 2 3003721 -1 -1 1 2333 2 1509012 -1 -1 3 1236 1235 1234 1 1001235 5678 9012 0 1 502222 1236 2468 2 6661 6662 1 3002333 -1 3721 3 6661 6662 6663 1 100

Sample Output:

38888 1 1.000 1000.0000001 15 0.600 100.0005551 4 0.750 100.000

两个代码，仅供参考：

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 using namespace std; 7 struct Node 8 { 9     int ID, nums = 0;10     double ss = 0.0, aa = 0.0;11 };12 int n;13 double num[10000] = { 0.0 }, area[10000] = { 0.0 };//房子数量//房子面积14 int mark[10000];//标记属于哪个家族15 vector
          
           
            >sets(10000);//记录每个家庭有多少人16 vector
            
             res;17 void combin(int my, int other)//合并家族18 {19 for (auto ptr = sets[other].begin(); ptr != sets[other].end(); ++ptr)//将其他成员进行同化20 {21 mark[*ptr] = my;22 sets[my].insert(*ptr);23 }24 sets[other].clear();//删除其他成员的标记25 }26 int main()27 { 28 cin >> n; 29 fill(mark, mark + 10000, -1);30 for (int i = 0; i < n; ++i)31 {32 int my, parent, k, child, nn, aa;33 cin >> my;34 if (mark[my] == -1)//还未标记是哪个家族35 {36 mark[my] = my;//就以自己为标记37 sets[mark[my]].insert(my);38 }39 for (int i = 0; i < 2; ++i)//添加父母40 {41 cin >> parent;42 if (parent == -1)43 continue;44 if (mark[parent] == -1)//家人未标记45 {46 mark[parent] = mark[my];47 sets[mark[my]].insert(parent);48 }49 else if (mark[parent] != -1 && mark[parent] != mark[my])//家人与自己标记不同50 combin(mark[my], mark[parent]);//将家人同化为自己标记51 }52 cin >> k;53 while (k--)54 {55 cin >> child;56 if (mark[child] == -1)//孩子未标记57 {58 mark[child] = mark[my];59 sets[mark[my]].insert(child);60 }61 else if (mark[child] != -1 && mark[child] != mark[my])//孩子与自己标记不同62 combin(mark[my], mark[child]);//将孩子同化为自己标记63 }64 cin >> nn >> aa;65 num[my] = nn;66 area[my] = aa;67 }68 for (int i = 0; i < sets.size(); ++i)69 {70 if (sets[i].empty())71 continue;72 Node* temp = new Node;73 temp->ID = *sets[i].begin();74 temp->nums = sets[i].size();75 for (auto ptr = sets[i].begin(); ptr != sets[i].end(); ++ptr)76 {77 temp->aa += area[*ptr];78 temp->ss += num[*ptr];79 }80 temp->ss /= temp->nums;81 temp->aa /= temp->nums;82 res.push_back(temp);83 }84 sort(res.begin(), res.end(), [](Node*a, Node*b) {85 if (abs(a->aa - b->aa) < 0.0001)86 return a->ID < b->ID;87 else88 return a->aa > b->aa; });89 cout << res.size() << endl;90 for (auto v : res)91 printf("%04d %d %0.3f %0.3f\n", v->ID, v->nums, v->ss, v->aa);92 return 0;93 }
            
           
          
         
        
       
      
     

1 #include
     
       2 using namespace std; 3 struct Estate{
   //存储集合最小id、集合结点个数num、集合总sets、集合总area 4     int id,num=0; 5     double sets=0.0,area=0.0; 6 }; 7 bool cmp(const Estate&e1,const Estate&e2){
   //比较函数 8     return e1.area!=e2.area?e1.area>e2.area:e1.id
      
       >idEstate;24 unordered_map
       
        familyEstate;25 int main(){26     int N;27     scanf("%d",&N);28     iota(father,father+10005,0);29     while(N--){
   //读取数据并合并相关集合30         int id,f,m,k,a;31         scanf("%d%d%d%d",&id,&f,&m,&k);32         if(f!=-1)33             unionSet(id,f);34         if(m!=-1)35             unionSet(id,m);36         while(k--){37             scanf("%d",&a);38             unionSet(id,a);39         }40         scanf("%lf%lf",&idEstate[id].first,&idEstate[id].second);//记录id和对应的sets、area41     }42     for(int i=0;i<10005;++i){
   //遍历并查集数组43         int temp=findFather(i);44         if(temp!=i)//根节点不等于本身45             ++familyEstate[temp].num;//递增根节点下集合的结点个数46         if(idEstate.find(i)!=idEstate.cend()){
   //累加集合的总sets、总area47             familyEstate[temp].sets+=idEstate[i].first;48             familyEstate[temp].area+=idEstate[i].second;49         }50     }51     vector
        
         v;52     for(auto i=familyEstate.begin();i!=familyEstate.end();++i){
   //将familyEstate中的值搬迁到vector中,并更新相关信息53         (i->second).id=i->first;54         ++(i->second).num;//集合下结点个数没有统计根节点，所以要+155         (i->second).sets/=(i->second).num;56         (i->second).area/=(i->second).num;57         v.push_back(i->second);58     }59     sort(v.begin(),v.end(),cmp);//排序60     printf("%d\n",v.size());61     for(int i=0;i