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 Child1⋯Childk Mestate 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; Childi's are the ID
's of his/her children; Mestate 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
AVGsets AVGarea
where ID
is the smallest ID in the family; M
is the total number of family members; AVGsets is the average number of sets of their real estate; and AVGarea 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~
#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;
struct node1 {
int ID, fID, mID, k, Me, area;
vector <int> child;
};
struct node2 {
int sumM = 0, sumSets = 0, sumArea = 0;
};
struct node3 {
int ID, M;
float AVGs, AVGa;
};
vector <int> S(10000);
vector <bool> visit(10000, false);
void initialize();//对集合S进行初始化
bool cmp(node3 &a, node3 &b);//排序规则
int getNum(node1 &a);//获取家庭成员的数量
int find(int X);
void unionSet(int X, int Y);
int main()
{
int N;
scanf("%d", &N);
vector <node1> data(N);
unordered_map <int, node2> mp;
initialize();
for (int i = 0; i < N; i++) {
node1 tmp;
scanf("%d%d%d%d", &tmp.ID, &tmp.fID, &tmp.mID, &tmp.k);
if (tmp.fID != -1)
unionSet(tmp.ID, tmp.fID);
if (tmp.mID != -1)
unionSet(tmp.ID, tmp.mID);
if (tmp.k != 0) {
for (int j = 0; j < tmp.k; j++) {
int cID;
scanf("%d", &cID);
tmp.child.push_back(cID);
unionSet(tmp.ID, cID);
}
}
scanf("%d%d", &tmp.Me, &tmp.area);
data[i] = tmp;
}
for (int i = 0; i < N; i++) {
int ID = find(data[i].ID);
mp[ID].sumM += getNum(data[i]);
mp[ID].sumSets += data[i].Me;
mp[ID].sumArea += data[i].area;
}
vector <node3> ans;
for (auto it = mp.begin(); it != mp.end(); it++) {
int ID = it->first,
M = it->second.sumM;
float AVGs = it->second.sumSets*1.0 / M,
AVGa = it->second.sumArea*1.0 / M;
ans.push_back({ ID,M,AVGs,AVGa });
}
sort(ans.begin(), ans.end(), cmp);
printf("%d\n", ans.size());
for (int i = 0; i < ans.size(); i++)
printf("%04d %d %.3f %.3f\n", ans[i].ID, ans[i].M, ans[i].AVGs, ans[i].AVGa);
return 0;
}
int getNum(node1 &a) {
int n = 0;
if (!visit[a.ID]) {
n++;
visit[a.ID] = true;
}
if ((a.fID != -1 )&& (!visit[a.fID])) {
n++;
visit[a.fID] = true;
}
if ((a.mID != -1) && (!visit[a.mID])) {
n++;
visit[a.mID] = true;
}
if (a.k != 0) {
for (int i = 0; i < a.k; i++) {
if (!visit[a.child[i]]) {
n++;
visit[a.child[i]] = true;
}
}
}
return n;
}
void unionSet(int X, int Y) {
int rootX = find(X),
rootY = find(Y);
if (rootX > rootY)
S[rootX] = S[rootY];
else
S[rootY] = S[rootX];
}
int find(int X) {
if (S[X] == X) {
return X;
}
else {
return S[X] = find(S[X]);//递归地压缩路径
}
}
void initialize() {
for (int i = 0; i < 10000; i++)
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
}
bool cmp(node3 &a, node3 &b) {
if (a.AVGa != b.AVGa)
return a.AVGa > b.AVGa;
else
return a.ID < b.ID;
}