pat甲级1114

1114 Family Property（25 分）

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 (≤1000). 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≤k≤5) 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

注意ID没有说不等于0.因此判断时为大于等于0，而不是大于。否则测试点2、4会错误。

我用的深度优先搜索求联通分量来完成的这道题，另外用并查集好像也比较方便。

由于ID不连续，所以求连通分量之前先将所有ID放到vector里面，而且用邻接表存储图更节省空间和时间。

 #include <iostream>
 #include <vector>
 #include <algorithm>
 using namespace std;

 bool marked[];
 vector<vector<int>> G();
 int estate[], area[];

 struct family
 {
     int id = , size = ;
     double estate = 0.0, area = 0.0;
 };

 void connect(int id)
 {
     int t;
     cin >> t;
     if (t >= )
     {
         G[id].push_back(t);
         G[t].push_back(id);
     }
 }

 void dfs(int s, family& f)
 {
     marked[s] = true;
     f.size++;
     if (s < f.id) f.id = s;
     f.area += area[s];
     f.estate += estate[s];
     for (int w : G[s])
         if (!marked[w]) dfs(w, f);
 }

 bool cmp(family a, family b)
 {
     if (a.area != b.area) return a.area > b.area;
     else return a.id < b.id;
 }

 int main()
 {
     int N;
     cin >> N;
     int i, j, id, k;
     vector<int> allId;
     for (i = ; i < N; i++)
     {
         cin >> id;
         allId.push_back(id);
         for (j = ; j < ; j++)
             connect(id);
         cin >> k;
         for (j = ; j < k; j++)
             connect(id);
         cin >> estate[id] >> area[id];
     }
     vector<family> v;
     for (int id : allId)
     {
         if (!marked[id])
         {
             family f;
             dfs(id, f);
             f.area /= f.size;
             f.estate /= f.size;
             v.push_back(f);
         }
     }
     sort(v.begin(), v.end(), cmp);
     cout << v.size() << endl;
     for (i = ; i < v.size(); i++)
         printf("%04d %d %.3f %.3f\n", v[i].id, v[i].size, v[i].estate, v[i].area);
     return ;
 }