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 Child1⋯Childk Mestate 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; Childi's are the ID's of his/her children; Mestate 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 AVGsets AVGarea
where ID is the smallest ID in the family; M is the total number of family members; AVGsets is the average number of sets of their real estate; and AVGarea 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：
1 #include <iostream> 2 #include <vector> 3 #include <set> 4 #include <algorithm> 5 #include <cmath> 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<set<int>>sets(10000);//记录每个家庭有多少人 16 vector<Node*>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 else 88 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<bits/stdc++.h> 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<e2.id; 9 } 10 int father[10005];//并查集底层数组 11 int findFather(int x){//查找根节点 12 if(x==father[x]) 13 return x; 14 int temp=findFather(father[x]); 15 father[x]=temp; 16 return temp; 17 } 18 void unionSet(int a,int b){//合并两个集合 19 int ua=findFather(a),ub=findFather(b); 20 if(ua!=ub) 21 father[max(ua,ub)]=min(ua,ub);//以小的id作为根节点 22 } 23 unordered_map<int,pair<double,double>>idEstate; 24 unordered_map<int,Estate>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、area 41 } 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area 47 familyEstate[temp].sets+=idEstate[i].first; 48 familyEstate[temp].area+=idEstate[i].second; 49 } 50 } 51 vector<Estate>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+1 55 (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<v.size();++i) 62 printf("%04d %d %.3f %.3f\n",v[i].id,v[i].num,v[i].sets,v[i].area); 63 return 0; 64 }