1114. Family Property (25)

梦里梦外; 2022-05-28 13:50 142阅读 0赞

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 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; **Childi**'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

--------------------

题目大意：

代码：

#include<stdio.h>
    #include<algorithm>
    using namespace std;
    struct node
    {
        int id,fatherid,motherid,M_estate,Area;
        int childid[20];
    }family[1001];
    struct node1
    {
        int id,person;
        double M_estate,Area;
        int flag;
    }answer[10001];
    int F[10001];
    int visited[100001];
    bool cmp(struct node1 a,struct node1 b)
    {
        if(a.Area!=b.Area)
            return a.Area>b.Area;
        else
            return a.id<b.id;
    }
    int find(int x)
    {
        while(x!=F[x])
        {
            x=F[x];
        }
        return x;
    }
    void Union(int x,int y)
    {
        int fx=find(x);
        int fy=find(y);
        if(fx!=fy)
        {
            if(fx>fy)
            {
                F[fx]=fy;
            }
            else
            {
               F[fy]=fx;
            }
        }
    }
    int main()
    {
        int i,j,n,m,k,t,cnt=0;
        for(i=0;i<10000;i++)
            F[i]=i;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d %d %d %d",&family[i].id,&family[i].fatherid,&family[i].motherid,&k);
            visited[family[i].id]=1;
            if(family[i].fatherid!=-1)
            {
                visited[family[i].fatherid]=1;
                Union(family[i].id,family[i].fatherid);
            }
            if(family[i].motherid!=-1)
            {
                visited[family[i].motherid]=1;
                Union(family[i].id,family[i].motherid);
            }
            for(j=0;j<k;j++)
            {
                scanf("%d",&family[i].childid[j]);
                visited[family[i].childid[j]]=1;
                Union(family[i].id,family[i].childid[j]);
            }
            scanf("%d %d",&family[i].M_estate,&family[i].Area);
        }
        for(i=0;i<n;i++)
        {
            int id=find(family[i].id);
            answer[id].id=id;
            answer[id].M_estate+=family[i].M_estate;
            answer[id].Area+=family[i].Area;
            answer[id].flag=1;
        }
        for(i=0;i<10000;i++)
        {
            if(visited[i]==1)
                answer[find(i)].person++;
            if(answer[i].flag==1)
                cnt++;
        }
        for(i=0;i<10000;i++) {
            if(answer[i].flag==1) {
                answer[i].M_estate = (double)(answer[i].M_estate * 1.0 / answer[i].person);
                answer[i].Area = (double)(answer[i].Area * 1.0 / answer[i].person);
            }
        }
        sort(answer,answer+10000,cmp);
        printf("%d\n", cnt);
        for(i=0;i<cnt;i++)
            printf("%04d %d %.3f %.3f\n",answer[i].id,answer[i].person,answer[i].M_estate,answer[i].Area);
        return 0;
    }