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L2-007 Family property (use of vector, set, map)

2022-07-30 15:42:00 【The real question is OK】

Given each person's family members and the real estate in their own name, please count the population, per capita real estate area and real estate units of each family.

input format:

Enter the first line to give a positive integer N (≤1000), followed by N lines, each of which gives a person's property in the following format:

No. Parent k Child 1 ... Child k Number of Property Units Total Area

Where number is a 4-digit number unique to each person; parent and parent are the person corresponding to the number respectively's parent's number (-1 if deceased); k (0≤k≤5) is that person's child'snumber; child i is the number of its child.

Output format:

First output the number of families on the first line (all people who are related belong to the same family).The information for each household is then output in the following format:

Minimum number of family members Household population Number of property units per capita Property area per capita

The per capita value requires 3 decimal places.The household information is first output in descending order of per capita area, and if there are parallels, it is output in ascending order of member number.

Input sample:

106666 5551 5552 1 7777 1 1001234 5678 9012 1 0002 2 3008888 -1 -1 0 1 10002468 0001 0004 1 2222 1 5007777 6666 -1 0 2 3003721 -1 -1 1 2333 2 1509012 -1 -1 3 1236 1235 1234 1 1001235 5678 9012 0 1 502222 1236 2468 2 6661 6662 1 3002333 -1 3721 3 6661 6662 6663 1 100

Example output:

38888 1 1.000 1000.0000001 15 0.600 100.0005551 4 0.750 100.000
#include #include #include #include #include using namespace std;const int N = 1e4 + 10;// Store the basic information to start typingstruct node{int id, house;double area;};// output family structurestruct family{int id, number;double house, area;};// custom comparison functionbool cmp(family a, family b){if (a.area != b.area) return a.area > b.area;return a.id < b.id;}int n, x, p[N];set s;// store the number of each person, the default ascending ordermap nodes; // record the property information of the person// And check the template, findint find(int x){if (p[x] != x) p[x] = find(p[x]);return p[x];}// And check template, mergevoid add(int x, int y){if (find(x) != find(y)) p[find(y)] = find(x);}int main(){scanf("%d", &n);for (int i = 0; i < N; i ++) p[i] = i;for (int i = 0; i < n; i++){// Record No. No. Parent k Child 1 ... Child k Number of property units Total areaint id, p1, p2, k, house, id1;double area; // total areacin >> id >> p1 >> p2 >> k;s.insert(id);if (p1 != -1) s.insert(p1), add(id, p1);if (p2 != -1) s.insert(p2), add(id, p2);for (int j = 0; j < k; j++)cin >> id1, s.insert(id1), add(id, id1);cin >> house >> area;nodes[id] = {id, house, area}; // record the person's family information}set s1; // Record the family number (it can be understood as recording the boss of each family)map> m; // record family number and member numberfor (auto it = s.begin(); it != s.end(); it++){int fa = find(*it);m[fa].push_back(*it);s1.insert(fa);}cout << s1.size() << endl;vector v1;for (auto it = s1.begin(); it != s1.end(); it++){int fa = *it;double house = 0, area = 0;// Calculate and summarize the members of each family and their properties and put them in the family structurefor (auto tmp = 0; tmp < m[fa].size(); tmp++)house += nodes[m[fa][tmp]].house, area += nodes[m[fa][tmp]].area;v1.push_back(family{ m[fa][0], int(m[fa].size()), house * 1.0 / (int)m[fa].size(), area * 1.0 / (int)m[fa].size()});}// put in ordersort(v1.begin(), v1.end(), cmp);for (int i = 0; i < v1.size(); i++)printf("%04d %d %.3lf %.3lf\n", v1[i].id, v1[i].number, v1[i].house, v1[i].area);return 0;}

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