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.. M 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; 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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出n个人以及他们的亲属、财产、房产面积，要求统计家族总人数、人均财产、人均面积。按人均面积递减排序，如果相同按id递增排序。

分析

一开始想用联通分量做，不是很方便。后来才想到了并查集，还是不熟练。

输入时处理家族关系。然后根据出现过的id找到他的家族，统计家族信息。然后根据出现过的家族id，存入一个新的vector。排序输出。

代码

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <deque>
#include <queue>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int n;
struct person{
    int id;
    int father;
    int mother;
    vector<int> childs;
    double estate;
    double area;
};
struct fam{
    int id;
    int num;
    double estate;
    double area;
};
vector<fam> famliy(10000);
int parent[10000];
int showup[10000],fshowup[10000];
person peo[10100];

int findpar(int a){
    while(parent[a]!=a){
        a=parent[a];
    }
    return a;
}
int makeunion(int a,int b){
    int pa=findpar(a);
    int pb=findpar(b);
    if(pa<pb)
        parent[pb]=pa;
    else
        parent[pa]=pb;
    return pa<pb?pa:pb;
}
bool cmp(fam f1,fam f2){
    if(f1.area/f1.num!=f2.area/f2.num)
        return f1.area/f1.num>f2.area/f2.num;
    else
        return f1.id<f2.id;
}
int main() {
    cin>>n;
    int id,father,mother,k,c;
    for(int i=0;i<10000;i++)
        parent[i]=i;
    for(int i=0;i<n;i++){
        cin>>id>>father>>mother>>k;
        peo[i].id=id;
        showup[id]=1;
        if(father!=-1) {
            makeunion(id, father);
            showup[father]=1;
        }
        if(mother!=-1) {
            showup[mother]=1;
            makeunion(id, mother);
        }
        for(int j=0;j<k;j++){
            cin>>c;
            showup[c]=1;
            makeunion(id,c);
        }
        cin>>peo[id].estate>>peo[id].area;
    }
    for(int i=0;i<10000;i++){
        if(showup[i]) {
            int fid = findpar(i);
            fshowup[fid] = 1;
            famliy[fid].id = fid;
            famliy[fid].num++;
            famliy[fid].estate += peo[i].estate;
            famliy[fid].area += peo[i].area;
        }
    }
    
    int sum=0;
    for(int i:fshowup)
        sum+=i;
    vector<fam> res;
    for(int i=0;i<10000;i++)
        if(fshowup[i])
            res.push_back(famliy[i]);
    
    sort(res.begin(),res.end(),cmp);
    cout<<sum<<endl;
    for(int i=0;i<sum;i++){
            double es=res[i].estate;
            double ar=res[i].area;
            printf("%04d %d %.3f %.3f\n",res[i].id,res[i].num,es/res[i].num,ar/res[i].num);
    }
    return 0;
}