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:

  1 /*
  2 Data: 2019-06-23 15:15:59
  3 Problem: PAT_A1114#Family Property
  4 AC: 42:40
  5 
  6 题目大意：
  7 统计家庭拥有的财产
  8 输入：
  9 第一行给出，户主人数N<=1000
 10 接下来N行，my_id,dad_id,mom_id,m,kids,房产数，总面积
 11 输出：
 12 第一行给出，家庭数N
 13 接来下N行，min_id,成员数,平均房产数，平均面积（三位小数）
 14 平均面积递减，min_id递增
 15 
 16 基本思路：
 17 输入时按id做并查集，min_id作为父结点，存储户主信息
 18 遍历户主，统计各家庭信息
 19 排序
 20 */
 21 #include<cstdio>
 22 #include<set>
 23 #include<vector>
 24 #include<algorithm>
 25 using namespace std;
 26 const int M=1e5+10;
 27 int fa[M];
 28 struct node
 29 {
 30     int id;
 31     set<int> member;
 32     double sets;
 33     double area;
 34 }info[M];
 35 
 36 int Father(int v)
 37 {
 38     int x=v,s;
 39     while(fa[v] != v)
 40         v = fa[v];
 41     while(fa[x] != x)
 42     {
 43         s = fa[x];
 44         fa[x] = v;
 45         x = s;
 46     }
 47     return v;
 48 }
 49 
 50 void Union(int v1, int v2)
 51 {
 52     int f1 = Father(v1);
 53     int f2 = Father(v2);
 54     if(f1 < f2)
 55         fa[f2] = f1;
 56     else
 57         fa[f1] = f2;
 58 }
 59 
 60 bool cmp(node a, node b)
 61 {
 62     int s1=a.member.size();
 63     int s2=b.member.size();
 64     if(a.area/s1 != b.area/s2)
 65         return a.area/s1 > b.area/s2;
 66     else
 67         return a.id < b.id;
 68 }
 69 
 70 int main()
 71 {
 72 #ifdef ONLINE_JUDGE
 73 #else
 74     freopen("Test.txt", "r", stdin);
 75 #endif // ONLINE_JUDGE
 76 
 77     for(int i=0; i<M; i++)
 78     {
 79         fa[i]=i;
 80         info[i].id=i;
 81         info[i].sets=0;
 82         info[i].area=0;
 83     }
 84 
 85     int n,my,dad,mom,m,kid;
 86     int input[M];
 87     scanf("%d", &n);
 88     for(int i=0; i<n; i++)
 89     {
 90         scanf("%d%d%d%d", &my,&dad,&mom,&m);
 91         input[i]=my;
 92         info[my].member.insert(my);
 93         if(dad != -1)
 94         {
 95             Union(my,dad);
 96             info[my].member.insert(dad);
 97         }
 98         if(mom != -1)
 99         {
100             Union(my,mom);
101             info[my].member.insert(mom);
102         }
103         for(int j=0; j<m; j++)
104         {
105             scanf("%d", &kid);
106             Union(my,kid);
107             info[my].member.insert(kid);
108         }
109         scanf("%lf%lf", &info[my].sets, &info[my].area);
110     }
111     set<int> family;
112     for(int i=0; i<n; i++)
113     {
114         my = input[i];
115         int f = Father(my);
116         family.insert(f);
117         if(f == my)
118             continue;
119         info[f].sets += info[my].sets;
120         info[f].area += info[my].area;
121         for(auto it=info[my].member.begin(); it!=info[my].member.end(); it++)
122             info[f].member.insert(*it);
123     }
124     vector<node> ans;
125     for(auto it=family.begin(); it!=family.end(); it++)
126         ans.push_back(info[*it]);
127     sort(ans.begin(),ans.end(),cmp);
128     printf("%d\n", ans.size());
129     for(int i=0; i<ans.size(); i++){
130         int sum = ans[i].member.size();
131         printf("%04d %d %.3f %.3f\n", ans[i].id,sum,ans[i].sets/sum,ans[i].area/sum);
132     }
133 
134     return 0;
135 }

转载于:https://www.cnblogs.com/blue-lin/p/11073307.html