PAT (Advanced Level)-1114 Family Property（dfs连通域）

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本文介绍了一种用于统计家族成员及家庭财产平均面积和数量的算法。通过深度优先搜索（DFS），该算法能够准确地计算出每个家族的大小、成员平均拥有的房产数量和面积。输入包括各家庭成员关系及个人名下房产信息，输出则为各家族的统计数据。

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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

思路：用dfs直接找连通域累加人数和地产信息并维护一个id 保持最小

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn(10005);
int set[maxn],area[maxn];
int book[maxn];
vector<int>v[maxn];

int sum,id;
double s,a;
struct q
{
	int id,ppl;
	double set,area;
	q(int idx,int sum,double s,double a):id(idx),ppl(sum),set(s),area(a){}
};
bool cmp(q x,q y)
{
	return x.area==y.area?x.id<y.id:x.area>y.area;
}
void dfs(int x)
{
	id=min(id,x);
	sum++;
	s+=set[x];
	a+=area[x];
	book[x]=0;
	for(int i=0;i<v[x].size();i++)
	{
		if(book[v[x][i]])dfs(v[x][i]);
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		int idx,fa,ma,k,child;
		scanf("%d %d %d",&idx,&fa,&ma);
		book[idx]=1;
		if(fa!=-1)
		{
			book[fa]=1;
			v[idx].push_back(fa);
			v[fa].push_back(idx);
		}
		if(ma!=-1)
		{
			book[ma]=1;
			v[idx].push_back(ma);
			v[ma].push_back(idx);
		}
		scanf("%d",&k);
		while(k--)
		{
			scanf("%d",&child);
			book[child]=1;
			v[idx].push_back(child);
			v[child].push_back(idx);
		}
		scanf("%d %d",&set[idx],&area[idx]);
	}
	vector<q>res;
	for(int i=0;i<10000;i++)
	{
		if(book[i])
		{
			s=a=0;
			id=10000;
			sum=0;
			dfs(i);
			res.push_back(q(id,sum,s/sum,a/sum));
		}
	}
	sort(res.begin(),res.end(),cmp);
	cout<<res.size()<<endl;
	for(int i=0;i<res.size();i++)
	{
		printf("%04d %d %.3lf %.3lf\n",res[i].id,res[i].ppl,res[i].set,res[i].area);
	}
}