PAT (Advanced Level) Practice 1114 Family Property (25分) （并查集+标记）

1.题目

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

2.题目分析

没超时没爆内存很是欣慰

1.使用并查集来寻找最小的ID

2.node2用来存放已给出节点的set以及area（有的节点的父母未给出set，area，这里只存放第一列ID以及其对应的set、area）

然后使用list2进行标记（开始想着遍历寻找可能会超时，就使用了标记）

3.unordered_set<int>ids;用来存放所有出现过的ID（-1除外），包括后面的父母、孩子

4.unordered_set<int>unique;用来存放所有的祖先节点

5.node用来存放最后要输出的一个总家庭的人数、平均set、评价area，放入set<node, cmp>out;输出 

注意：代码中：if (find(*it2) == *it)//不能是father[*it2]==*it

即使经过路径压缩，最后每个节点的父亲都不一定是最后的祖先！！！！！所以在判断节点是不是祖先的时候不能直接使用father【】，而要使用find（）！！！

3.代码

#include<iostream>
#include<set>
#include<unordered_set>
using namespace std;
int father[10001];
struct node
{
	int id;
	int mcount;
	double avsets;
	double avarea;
};
struct node2
{
	int id;
	int mcount;
	double area;
}list2[10001];
struct cmp
{
	bool operator () (const node &a, const node &b)
	{
		if (a.avarea == b.avarea)return a.id < b.id;
		return a.avarea > b.avarea;
	}
};
int find(int x)
{
	if (father[x] == x)return x;
	return father[x]=find(father[x]);
}

void unions(int x, int y)
{
	int a = find(x);
	int b =find(y);
	if (a == b)return;
	if (a > b)father[a] = b;
	if (a < b)father[b] = a;
}
int main()
{
	int n,a,b;
	scanf("%d", &n);
	for (int i = 0; i < 10001; i++)
		father[i] = i;
	unordered_set<int>ids;
	unordered_set<int>unique;
	set<node, cmp>out;
	for (int i = 0; i < n; i++)
	{
		int k, c, mark, m;
		double areas;
		scanf("%d %d %d", &mark, &a, &b);
		ids.insert(mark);
		scanf("%d", &k);
		for (int j = 0; j < k; j++)
		{
			scanf("%d", &c);
			ids.insert(c);
			unions(mark, c);
		}
		scanf("%d %lf", &m, &areas);
		list2[mark].id = mark; list2[mark].mcount = m; list2[mark].area = areas;
		if (a != -1) { ids.insert(a); unions(mark, a); }
		if (b != -1) { ids.insert(b); unions(mark, b); }
	}

	for (auto it = ids.begin(); it != ids.end(); it++)
	{
		int aaaaa = father[*it];
		if (*it == father[*it])
			unique.insert(*it);
	}
	for (auto it = unique.begin(); it != unique.end(); it++)
	{
		int m = 0; double sumset = 0, sumarea = 0;
		for (auto it2 = ids.begin(); it2 != ids.end(); it2++)
		{
			if (find(*it2) == *it)//这里注意不能是father[*it2]==*it
			{
				m++;
				sumset += list2[*it2].mcount;
				sumarea += list2[*it2].area;
			}
		}
		out.insert(node{*it,m,sumset/m,sumarea/m});
	}

	printf("%d\n", out.size());
	for (auto it = out.begin(); it != out.end(); it++)
	{
		printf("%04d %d %.3f %.3f\n", it ->id, it->mcount, it->avsets, it->avarea);
	}

}