题目地址

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

#include <iostream>
#include <vector>
#include<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=10010,inf=100000000;
int n;double to_mes=0,to_area=0;int min_id,cnt=0;
struct node{
    int id;
    double area,mestate;
    vector<int>v;
}s[maxn];
bool vis[maxn]={0};
struct ans_node{
    int id,cnt;double ave_mes,ave_area;
};
vector<ans_node>ans;
void dfs(int idx){
    if(vis[idx]) return;
    if(idx<min_id) min_id=idx;
    if(s[idx].mestate) to_mes+=s[idx].mestate;
    if(s[idx].area)to_area+=s[idx].area;
    cnt++;vis[idx]=1;
    for(int i=0;i<s[idx].v.size();i++){
        if(!vis[s[idx].v[i]]) dfs(s[idx].v[i]);
    }
}
bool cmp(ans_node a,ans_node b){
    if(a.ave_area!=b.ave_area) return a.ave_area>b.ave_area;
    else if(a.id!=b.id) return a.id<b.id;
}
int main(){
    cin>>n;int id,fa,ma,k,chi,temp[n+1],temp_cnt=0;
    for(int i=0;i<n;i++){
        cin>>id;s[id].id=id;temp[temp_cnt++]=id;
        cin>>fa>>ma>>k;
        if(fa!=-1){
            s[id].v.push_back(fa);s[fa].v.push_back(id);
        }
        if(ma!=-1){
            s[id].v.push_back(ma);s[ma].v.push_back(id);
        }
        for(int i=0;i<k;i++){
            cin>>chi;s[id].v.push_back(chi);s[chi].v.push_back(id);
        }
        cin>>s[id].mestate>>s[id].area;
    }int pro_cnt=0;
    for(int i=0;i<n;i++){
        if(vis[temp[i]])continue;
        min_id=inf;cnt=0;to_area=0;to_mes=0;
        dfs(temp[i]);
        pro_cnt++;
        ans.push_back({min_id,cnt,to_mes/cnt,to_area/cnt});
    }
    cout<<pro_cnt<<endl;
    sort(ans.begin(),ans.end(),cmp);
    for(int i=0;i<ans.size();i++){
        printf("%04d %d %.3lf %.3lf\n",ans[i].id,ans[i].cnt,ans[i].ave_mes,ans[i].ave_area);
    }
}