PAT甲级——1114 Family Property (并查集)


此文章同步发布在我的CSDN上https://blog.csdn.net/weixin_44385565/article/details/89930332

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 (≤). 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

题目大意：根据给出的数据分类出各个家庭，然后输出家庭个数；接下来输出每个家庭的成员ID（只需输出家庭成员中最小的ID）、成员数量、平均房产数量、平均房产面积，输出顺序为平均房产面积降序、ID升序。

思路：用并查集S给家庭分类，为了方便找出最小的家庭成员ID，对并查集的操作稍作修改，集合元素初始化为对应的下标，若S[X]==X，则X为根，unionSet()函数将大根集合并到小根集合里；家庭成员计数时需要对已经访问过的ID进行标记，防止重复计数，用visit数组标记每个ID。

总结：其实我是比较害怕这类问题的，看起来简单，可是信息一多思考起来就容易混乱，需要很细心很耐心地分析，用不到什么算法知识却超级消耗时间~

  1 #include <iostream>
  2 #include <vector>
  3 #include <unordered_map>
  4 #include <algorithm>
  5 using namespace std;
  6 struct node1 {
  7     int ID, fID, mID, k, Me, area;
  8     vector <int> child;
  9 };
 10 struct node2 {
 11     int sumM = 0, sumSets = 0, sumArea = 0;
 12 };
 13 struct node3 {
 14     int ID, M;
 15     float AVGs, AVGa;
 16 };
 17 vector <int> S(10000);
 18 vector <bool> visit(10000, false);
 19 void initialize();//对集合S进行初始化
 20 bool cmp(node3 &a, node3 &b);//排序规则
 21 int getNum(node1 &a);//获取家庭成员的数量
 22 int find(int X);
 23 void unionSet(int X, int Y);
 24 int main()
 25 {
 26     int N;
 27     scanf("%d", &N);
 28     vector <node1> data(N);
 29     unordered_map <int, node2> mp;
 30     initialize();
 31     for (int i = 0; i < N; i++) {
 32         node1 tmp;
 33         scanf("%d%d%d%d", &tmp.ID, &tmp.fID, &tmp.mID, &tmp.k);
 34         if (tmp.fID != -1)
 35             unionSet(tmp.ID, tmp.fID);
 36         if (tmp.mID != -1)
 37             unionSet(tmp.ID, tmp.mID);
 38         if (tmp.k != 0) {
 39             for (int j = 0; j < tmp.k; j++) {
 40                 int cID;
 41                 scanf("%d", &cID);
 42                 tmp.child.push_back(cID);
 43                 unionSet(tmp.ID, cID);
 44             }
 45         }
 46         scanf("%d%d", &tmp.Me, &tmp.area);
 47         data[i] = tmp;
 48     }
 49     for (int i = 0; i < N; i++) {
 50         int ID = find(data[i].ID);
 51         mp[ID].sumM += getNum(data[i]);
 52         mp[ID].sumSets += data[i].Me;
 53         mp[ID].sumArea += data[i].area;
 54     }
 55     vector <node3> ans;
 56     for (auto it = mp.begin(); it != mp.end(); it++) {
 57         int ID = it->first,
 58             M = it->second.sumM;
 59         float AVGs = it->second.sumSets*1.0 / M,
 60             AVGa = it->second.sumArea*1.0 / M;
 61         ans.push_back({ ID,M,AVGs,AVGa });
 62     }
 63     sort(ans.begin(), ans.end(), cmp);
 64     printf("%d\n", ans.size());
 65     for (int i = 0; i < ans.size(); i++)
 66         printf("%04d %d %.3f %.3f\n", ans[i].ID, ans[i].M, ans[i].AVGs, ans[i].AVGa);
 67     return 0;
 68 }
 69 int getNum(node1 &a) {
 70     int n = 0;
 71     if (!visit[a.ID]) {
 72         n++;
 73         visit[a.ID] = true;
 74     }
 75     if ((a.fID != -1 )&& (!visit[a.fID])) {
 76         n++;
 77         visit[a.fID] = true;
 78     }
 79     if ((a.mID != -1) && (!visit[a.mID])) {
 80         n++;
 81         visit[a.mID] = true;
 82     }
 83     if (a.k != 0) {
 84         for (int i = 0; i < a.k; i++) {
 85             if (!visit[a.child[i]]) {
 86                 n++;
 87                 visit[a.child[i]] = true;
 88             }
 89         }
 90     }
 91     return n;
 92 }
 93 void unionSet(int X, int Y) {
 94     int rootX = find(X),
 95         rootY = find(Y);
 96     if (rootX > rootY)
 97         S[rootX] = S[rootY];
 98     else 
 99         S[rootY] = S[rootX];
100 }
101 int find(int X) {
102     if (S[X] == X) {
103         return X;
104     }
105     else {
106         return S[X] = find(S[X]);//递归地压缩路径
107     }
108 }
109 void initialize() {
110     for (int i = 0; i < 10000; i++) 
111         S[i] = i;//将集合元素初始化为对应的下标，每个集合的根节点就是当前家庭的最小ID
112 }
113 bool cmp(node3 &a, node3 &b) {
114     if (a.AVGa != b.AVGa) 
115         return a.AVGa > b.AVGa;
116     else 
117         return a.ID < b.ID;
118 }

 

 
posted @ 2019-05-07 22:02  大众名字重名太多  阅读(226)  评论(0)  编辑  收藏  举报