PAT甲级——A1114 Family Property【25】

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:

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

两个代码，仅供参考：

 #include <iostream>
 #include <vector>
 #include <set>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 struct Node
 {
     int ID, nums = ;
     double ss = 0.0, aa = 0.0;
 };
 int n;
 double num[] = { 0.0 }, area[] = { 0.0 };//房子数量//房子面积
 int mark[];//标记属于哪个家族
 vector<set<int>>sets();//记录每个家庭有多少人
 vector<Node*>res;
 void combin(int my, int other)//合并家族
 {
     for (auto ptr = sets[other].begin(); ptr != sets[other].end(); ++ptr)//将其他成员进行同化
     {
         mark[*ptr] = my;
         sets[my].insert(*ptr);
     }
     sets[other].clear();//删除其他成员的标记
 }
 int main()
 {
     cin >> n;
     fill(mark, mark + , -);
     for (int i = ; i < n; ++i)
     {
         int my, parent, k, child, nn, aa;
         cin >> my;
         if (mark[my] == -)//还未标记是哪个家族
         {
             mark[my] = my;//就以自己为标记
             sets[mark[my]].insert(my);
         }
         for (int i = ; i < ; ++i)//添加父母
         {
             cin >> parent;
             if (parent == -)
                 continue;
             if (mark[parent] == -)//家人未标记
             {
                 mark[parent] = mark[my];
                 sets[mark[my]].insert(parent);
             }
             else if (mark[parent] != - && mark[parent] != mark[my])//家人与自己标记不同
                 combin(mark[my], mark[parent]);//将家人同化为自己标记
         }
         cin >> k;
         while (k--)
         {
             cin >> child;
             if (mark[child] == -)//孩子未标记
             {
                 mark[child] = mark[my];
                 sets[mark[my]].insert(child);
             }
             else if (mark[child] != - && mark[child] != mark[my])//孩子与自己标记不同
                 combin(mark[my], mark[child]);//将孩子同化为自己标记
         }
         cin >> nn >> aa;
         num[my] = nn;
         area[my] = aa;
     }
     for (int i = ; i < sets.size(); ++i)
     {
         if (sets[i].empty())
             continue;
         Node* temp = new Node;
         temp->ID = *sets[i].begin();
         temp->nums = sets[i].size();
         for (auto ptr = sets[i].begin(); ptr != sets[i].end(); ++ptr)
         {
             temp->aa += area[*ptr];
             temp->ss += num[*ptr];
         }
         temp->ss /= temp->nums;
         temp->aa /= temp->nums;
         res.push_back(temp);
     }
     sort(res.begin(), res.end(), [](Node*a, Node*b) {
         if (abs(a->aa - b->aa) < 0.0001)
             return a->ID < b->ID;
         else
             return a->aa > b->aa; });
     cout << res.size() << endl;
     for (auto v : res)
         printf("%04d %d %0.3f %0.3f\n", v->ID, v->nums, v->ss, v->aa);
     return ;
 }

 #include<bits/stdc++.h>
 using namespace std;
 struct Estate{//存储集合最小id、集合结点个数num、集合总sets、集合总area
     int id,num=;
     double sets=0.0,area=0.0;
 };
 bool cmp(const Estate&e1,const Estate&e2){//比较函数
     return e1.area!=e2.area?e1.area>e2.area:e1.id<e2.id;
 }
 int father[];//并查集底层数组
 int findFather(int x){//查找根节点
     if(x==father[x])
         return x;
     int temp=findFather(father[x]);
     father[x]=temp;
     return temp;
 }
 void unionSet(int a,int b){//合并两个集合
     int ua=findFather(a),ub=findFather(b);
     if(ua!=ub)
         father[max(ua,ub)]=min(ua,ub);//以小的id作为根节点
 }
 unordered_map<int,pair<double,double>>idEstate;
 unordered_map<int,Estate>familyEstate;
 int main(){
     int N;
     scanf("%d",&N);
     iota(father,father+,);
     while(N--){//读取数据并合并相关集合
         int id,f,m,k,a;
         scanf("%d%d%d%d",&id,&f,&m,&k);
         if(f!=-)
             unionSet(id,f);
         if(m!=-)
             unionSet(id,m);
         while(k--){
             scanf("%d",&a);
             unionSet(id,a);
         }
         scanf("%lf%lf",&idEstate[id].first,&idEstate[id].second);//记录id和对应的sets、area
     }
     for(int i=;i<;++i){//遍历并查集数组
         int temp=findFather(i);
         if(temp!=i)//根节点不等于本身
             ++familyEstate[temp].num;//递增根节点下集合的结点个数
         if(idEstate.find(i)!=idEstate.cend()){//累加集合的总sets、总area
             familyEstate[temp].sets+=idEstate[i].first;
             familyEstate[temp].area+=idEstate[i].second;
         }
     }
     vector<Estate>v;
     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
         (i->second).id=i->first;
         ++(i->second).num;//集合下结点个数没有统计根节点，所以要+1
         (i->second).sets/=(i->second).num;
         (i->second).area/=(i->second).num;
         v.push_back(i->second);
     }
     sort(v.begin(),v.end(),cmp);//排序
     printf("%d\n",v.size());
     for(int i=;i<v.size();++i)
         printf("%04d %d %.3f %.3f\n",v[i].id,v[i].num,v[i].sets,v[i].area);
     return ;
 }