PAT_A1114#Family Property

Source:

PAT A1114 Family Property (25 分)

Description:

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

Keys:

• 并查集（Union-Find）

Attention:

• Union函数

Code:

 /*
 Data: 2019-06-23 15:15:59
 Problem: PAT_A1114#Family Property
 AC: 42:40

 题目大意：
 统计家庭拥有的财产
 输入：
 第一行给出，户主人数N<=1000
 接下来N行，my_id,dad_id,mom_id,m,kids,房产数，总面积
 输出：
 第一行给出，家庭数N
 接来下N行，min_id,成员数,平均房产数，平均面积（三位小数）
 平均面积递减，min_id递增

 基本思路：
 输入时按id做并查集，min_id作为父结点，存储户主信息
 遍历户主，统计各家庭信息
 排序
 */
 #include<cstdio>
 #include<set>
 #include<vector>
 #include<algorithm>
 using namespace std;
 const int M=1e5+;
 int fa[M];
 struct node
 {
     int id;
     set<int> member;
     double sets;
     double area;
 }info[M];

 int Father(int v)
 {
     int x=v,s;
     while(fa[v] != v)
         v = fa[v];
     while(fa[x] != x)
     {
         s = fa[x];
         fa[x] = v;
         x = s;
     }
     return v;
 }

 void Union(int v1, int v2)
 {
     int f1 = Father(v1);
     int f2 = Father(v2);
     if(f1 < f2)
         fa[f2] = f1;
     else
         fa[f1] = f2;
 }

 bool cmp(node a, node b)
 {
     int s1=a.member.size();
     int s2=b.member.size();
     if(a.area/s1 != b.area/s2)
         return a.area/s1 > b.area/s2;
     else
         return a.id < b.id;
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     for(int i=; i<M; i++)
     {
         fa[i]=i;
         info[i].id=i;
         info[i].sets=;
         info[i].area=;
     }

     int n,my,dad,mom,m,kid;
     int input[M];
     scanf("%d", &n);
     for(int i=; i<n; i++)
     {
         scanf("%d%d%d%d", &my,&dad,&mom,&m);
         input[i]=my;
         info[my].member.insert(my);
         if(dad != -)
         {
             Union(my,dad);
             info[my].member.insert(dad);
         }
         if(mom != -)
         {
             Union(my,mom);
             info[my].member.insert(mom);
         }
         for(int j=; j<m; j++)
         {
             scanf("%d", &kid);
             Union(my,kid);
             info[my].member.insert(kid);
         }
         scanf("%lf%lf", &info[my].sets, &info[my].area);
     }
     set<int> family;
     for(int i=; i<n; i++)
     {
         my = input[i];
         int f = Father(my);
         family.insert(f);
         if(f == my)
             continue;
         info[f].sets += info[my].sets;
         info[f].area += info[my].area;
         for(auto it=info[my].member.begin(); it!=info[my].member.end(); it++)
             info[f].member.insert(*it);
     }
     vector<node> ans;
     for(auto it=family.begin(); it!=family.end(); it++)
         ans.push_back(info[*it]);
     sort(ans.begin(),ans.end(),cmp);
     printf("%d\n", ans.size());
     for(int i=; i<ans.size(); i++){
         int sum = ans[i].member.size();
         printf("%04d %d %.3f %.3f\n", ans[i].id,sum,ans[i].sets/sum,ans[i].area/sum);
     }

     return ;
 }