1114 Family Property（25分）

题目翻译：

给定每个人的家庭成员和其自己名下的房产，请你统计出每个家庭的人口数、人均房产面积及房产套数。首先在第一行输出家庭个数（所有有亲属关系的人都属于同一个家庭）。随后按下列格式输出每个家庭的信息：家庭成员的最小编号 家庭人口数 人均房产套数 人均房产面积。其中人均值要求保留小数点后3位。家庭信息首先按人均面积降序输出，若有并列，则按成员编号的升序输出。

题解思路：

并查集或者dfs

代码：

dfs代码：

#include
using namespace std;
int N;
struct node {
	int set = 0, area = 0;
};
node value[10000];//存储结点的值
vector v[10000];//用于表示邻居关系
int visited[10000],rea[10000];//是否访问过，是否是关系网中的结点
struct results {//存储结果
	int min_id, num, sum1, sum2;
};
vector R;

int min_id = 0 , num = 0, S1 = 0, S2 = 0;//用于得到dfs的结果

void dfs(int curnode)
{
	visited[curnode] = 1;
	if (curnode < min_id)min_id = curnode;
	for (auto i : v[curnode])
	{
		if (visited[i] == 0)
		{
			dfs(i);
			num++, S1 += value[i].set, S2 += value[i].area;
		}
	}
}

bool comp(results r1, results r2)
{
	if (r1.sum2 / double(r1.num) > r2.sum2 / double(r2.num))
		return true;
	else if (r1.sum2 / double(r1.num) == r2.sum2 / double(r2.num))
		return r1.min_id < r2.min_id;
	else
		return false;
}

int main()
{
	cin >> N;
	int id1, id2, id3, k;
	for (int i = 0;i < N;i++)
	{
		cin >> id1 >> id2 >> id3;
		rea[id1] = 1;
		node n;
		if (id2 != -1)
		{
			rea[id2] = 1;
			v[id1].push_back(id2);
			v[id2].push_back(id1);
		}
		if (id3 != -1)
		{
			rea[id3] = 1;
			v[id1].push_back(id3);
			v[id3].push_back(id1);
		}
		cin >> k;
		while (k--)
		{
			int child;
			cin >> child;
			v[id1].push_back(child);
			v[child].push_back(id1);
		}
		cin >> value[id1].set >> value[id1].area;
	}
	for (int i = 0;i < 10000;i++)
	{
		if (!visited[i] && rea[i])
		{
			min_id = i, num = 1, S1 = value[i].set, S2 = value[i].area;//赋初值
			dfs(i);
			results temp;
			temp.min_id = min_id, temp.num = num, temp.sum1 = S1, temp.sum2 = S2;
			R.push_back(temp);
		}
	}
	sort(R.begin(), R.end(), comp);
	cout << R.size() << endl;
	for (auto i : R)
		cout <

并查集实现;

#include
using namespace std;
int N;
struct node {
	int set = 0, area = 0;
};
node value[10000];
vector v[10000];//邻居关系
int f[10000];//并查集存储结点的父亲节点
vector cv;//输入的最左侧的值
int rea[10000];//区分是否为原始点
struct results {
	int min_id = 10000, num = 0, sum1 = 0, sum2 = 0, tag = 0;
};
vector R;

bool comp(results r1, results r2)
{
	if (r1.sum2 / double(r1.num) > r2.sum2 / double(r2.num))
		return true;
	else if (r1.sum2 / double(r1.num) == r2.sum2 / double(r2.num))
		return r1.min_id < r2.min_id;
	else
		return false;
}

int find(int p)
{
	if (p != f[p]) f[p] = find(f[p]);
	return f[p];
}

int main()
{
	cin >> N;
	int id1, id2, id3, k;
	for (int i = 0;i < N;i++)
	{
		cin >> id1 >> id2 >> id3;
		cv.push_back(id1);
		rea[id1] = 1;
		if (id2 != -1)
		{
			v[id1].push_back(id2);
			rea[id2] = 1;
		}
		if (id3 != -1)
		{
			v[id1].push_back(id3);
			rea[id3] = 1;
		}
		cin >> k;
		while (k--)
		{
			int child;
			cin >> child;
			v[id1].push_back(child);
		}
		cin >> value[id1].set >> value[id1].area;
	}
	
	for (int i = 0;i < 10000;i++)
		f[i] = i;
	
	for (int i = 0;i < cv.size();i++)
	{
		for (int j = 0;j < v[cv[i]].size();j++)
		{
			int a = find(cv[i]), b = find(v[cv[i]][j]);
			if (a != b) f[b] = a;
		}
	}
	map R1;
	for (int i = 0;i < 10000;i++)
	{
		if (i < R1[find(i)].min_id)
			R1[find(i)].min_id = i;
		R1[find(i)].num++;
		R1[find(i)].sum1 += value[i].set;
		R1[find(i)].sum2 += value[i].area;
	}
	for (int i = 0;i < 10000;i++)
	{
		if ((R1[find(i)].num >= 2 || (R1[find(i)].num == 1 && rea[i]))&&R1[find(i)].tag==0)
		{
			R.push_back(R1[find(i)]);
			R1[find(i)].tag = 1;
		}
	}

	sort(R.begin(), R.end(), comp);
	cout << R.size() << endl;
	for (auto i : R)
		cout << setw(4) << setfill('0') << i.min_id << " " << i.num << " " << fixed << setprecision(3) << i.sum1 / double(i.num) << " " << fixed << setprecision(3) << i.sum2 / double(i.num) << endl;
}

坑点：

无