1114 Family Property （25 分）

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 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

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);
     }
     ;
 }