【PAT】1114. Family Property (25)

1114. Family Property (25)

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 ... 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

分析：用图（并查集）结构存储输入；用bfs遍历计算。

细节：图的边为无向边；

（以下代码比较丑，还是完整定义图结构比较迅速方便）

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <vector>
#include <functional>

#define rep(i,j,k) for(int i=j;i<=k;++i)
const int Max=10000;
using namespace std;

class node
{
	public:
		int id;
		int sets;
		int area;
		vector<int> family;
		node(){
			id=sets=area=0;
		};
};

struct res
{
	int minres;
	float set,are;
	int sum;
	friend bool operator < (const res& a, const res& b){
		return a.are<b.are || (a.are==b.are && a.minres>b.minres);
	}
};

vector<int> root;
node estate[Max];
int visited[Max];
priority_queue<res> result;

void create(int n)
{
	while (n--){
		node x;
		int id,f,m,k;
		scanf("%d %d %d %d",&id, &f, &m, &k);
		root.push_back(id);
		x.id=id;
		x.family.push_back(f);
		x.family.push_back(m);
		estate[f].family.push_back(id); // f==-1时仍然可以执行 
		estate[m].family.push_back(id); //
		while (k--){
			int c;
			scanf("%d",&c);
			x.family.push_back(c);
			estate[c].family.push_back(id);
		}
		int s,a;
		scanf("%d %d",&s, &a);
		x.sets=s;
		x.area=a;
		estate[id]=x;
	}
}

void bfs(int root)
{
	queue<int> q;
	int minres=10000;
	int sum=0;
	float set=0,are=0;
	q.push(root);
	visited[root]=1;
	int v;
	while(!q.empty()){
		v=q.front();
		if (v<minres) minres=v;
		++sum;
		set+=estate[v].sets;
		are+=estate[v].area;
		int size=estate[v].family.size();
		rep(i,0,size-1){
			if (estate[v].family[i]==-1) continue; 
			if (visited[estate[v].family[i]]!=1){
				visited[estate[v].family[i]]=1;
				q.push(estate[v].family[i]);
			}
		}
		q.pop();
	}
	res t;
	t.minres=minres;
	t.set=set/sum;
	t.are=are/sum;
	t.sum=sum;
	result.push(t);
}

void major()
{
	
	rep(i,1,root.size()){
		if (visited[root[i-1]]!=1){
			bfs(root[i-1]);
		}
	}
	cout<<result.size()<<endl;
	while (!result.empty()){
		res t=result.top();
		printf("%04d %d %.3f %.3f\n", t.minres, t.sum, t.set, t.are);
		result.pop();
	}
}


int main()
{
//	freopen("test.txt","r",stdin);
	int n;
	cin>>n;
	create(n);
	major(); 
	return 0;
}