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分析：纯粹的模拟题，按题目要求做即可，一开始一直过不了，后来加了个数组vis表示该人还存活，另外3、4、5测试点里有00000这个人要注意。
/**
* Copyright(c)
* All rights reserved.
* Author : Mered1th
* Date : 2019-02-27-13.50.01
* Description : A1114
*/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<string>
#include<unordered_set>
#include<map>
#include<vector>
#include<set>
using namespace std;
;
struct Family{
; //m是人口数
,Avgarea=;
,area=;//总套数，总面积
bool flag=false;
}family[maxn];
struct Node{
int id,f,m,k;
vector<int> child;
double Msets,Area;
}node[maxn];
int father[maxn];
int find_Father(int x){
int a=x;
while(x!=father[x]){
x=father[x];
}
while(a!=father[a]){
int z=a;
a=father[a];
father[z]=x;
}
return x;
}
void Union(int a,int b){
int faA=find_Father(a);
int faB=find_Father(b);
if(faA!=faB){
if(faA<faB){
father[faB]=faA;
}
else if(faA>faB){
father[faA]=faB;
}
}
}
bool cmp(Family a,Family b){
if(a.Avgarea!=b.Avgarea) return a.Avgarea>b.Avgarea;
else return a.id<b.id;
}
bool vis[maxn]={false};
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int n,k,id,child;
scanf("%d",&n);
;i<maxn;i++) father[i]=i;
;i<n;i++){
scanf("%d",&id);
scanf("%d%d%d",&node[id].f,&node[id].m,&k);
vis[id]=true;
node[id].id=id;
){
Union(id,node[id].f);
vis[node[id].f]=true;
}
){
Union(id,node[id].m);
vis[node[id].m]=true;
}
;j<k;j++){
scanf("%d",&child);
node[id].child.push_back(child);
vis[child]=true;
Union(id,child);
}
scanf("%lf%lf",&node[id].Msets,&node[id].Area);
}
;i<maxn;i++){
int faI=find_Father(node[i].id);
if(vis[faI]){
family[faI].id=faI;
family[faI].area+=node[i].Area;
family[faI].sets+=node[i].Msets;
family[faI].flag=true;
}
}
;
;i<maxn;i++){
if(vis[i]){
family[find_Father(i)].m++;
}
if(family[i].flag==true){
num++;
}
}
;i<maxn;i++){
){
family[i].Avgarea=family[i].area/family[i].m;
family[i].Avgsets=family[i].sets/family[i].m;
}
}
sort(family,family+maxn,cmp);
cout<<num<<endl;
;i<num;i++){
printf("%04d %d %.3f %.3f\n",family[i].id,family[i].m,family[i].Avgsets,family[i].Avgarea);
}
;
}