【刷题-PAT】A1114 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 child1 child2 ... childk 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:

\(ID M AVG_{sets} AVG_{area}\)

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：要将每个家庭的数据分离出来，使用并查集即可实现，数据处理上有两种方法：

1.先将录入的数据按照输入的形式存在数组中，录入的同时完成集合的合并，然后再从录入的数据中挑出需要的数据，最后对挑出的数据仓晒礼输出；

2.在录入的同时进行集合的合并，并将数据按照输出的格式存储，然后进行处理

#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<unordered_map>
#include<set>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;
const int nmax = 10100;
int fath[nmax];
void init(){
    for(int i = 0; i < nmax; ++i)fath[i] = i;
}
int findF(int x){
    int z = x;
    while(x != fath[x])x = fath[x];
    while(z != fath[z]){
        int temp = fath[z];
        fath[z] = x;
        z = temp;
    }
    return x;
}
void Union(int a, int b){
    int fa = findF(a), fb = findF(b);
    if(fa < fb)fath[fb] = fa;
    else if(fa > fb)fath[fa] = fb;
}
struct node{
    int id, people;
    double num, area;
    bool flag = false;
    bool operator < (node &a)const{
        return area != a.area ? area > a.area : id < a.id;
    }
}ans[nmax];
struct data{
    int id, fid, mid, k;
    int child[6];
    int num, area;
}v[nmax];
bool vis[nmax] = {false};
int main(){
    #ifdef ONLINE_JUDGE
    #else
    freopen("input.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    init();
    int N;
    scanf("%d", &N);
    //录入数据并合并，由于序号不连续，要标记哪些序号已经出现过
    for(int i = 0; i < N; ++i){
        int id, fid, mid, k;
        scanf("%d%d%d%d", &id, &fid, &mid, &k);
        v[id].id = id;
        v[id].fid = fid;
        v[id].mid = mid;
        v[id].k = k;
        vis[id] = true;
        if(fid != -1){
            Union(id, fid);
            vis[fid] = true;
        }
        if(mid != -1){
            Union(id, mid);
            vis[mid] = true;
        }
        for(int j = 0; j < k; ++j){
            scanf("%d", &v[id].child[j]);
            Union(id, v[id].child[j]);
            vis[v[id].child[j]] = true;
        }
        scanf("%d%d", &v[id].num, &v[id].area);
    }
    //统计并抽出需要的数据，用flag标记该根节点是否已经出现
    int cnt = 0;
    for(int i = 0; i < nmax; ++i){
        if(vis[i] == true){
            int index = findF(i);
            ans[index].id = index;
            ans[index].people++;
            ans[index].num += v[i].num;
            ans[index].area += v[i].area;
            if(ans[index].flag == false)cnt++;
            ans[index].flag = true;
        }
    }
    //处理
    for(int i = 0; i < nmax; ++i){
        if(ans[i].flag == true){
            ans[i].num /= ans[i].people;
            ans[i].area /= ans[i].people;
        }
    }
    //排序输出
    sort(ans, ans + nmax);
    printf("%d\n", cnt);
    for(int i = 0; i < cnt; ++i){
        printf("%04d %d %.3f %.3f\n", ans[i].id, ans[i].people, ans[i].num, ans[i].area);
    }
    return 0;
}

#include<cstdio>
#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
const int Nmax = 10000;
int father[Nmax];
bool vis[Nmax] = {false};
struct node{
    int id, peoNum;
    double Sav, Aav;
};
void init(){
    for(int i = 0; i < Nmax; ++i)father[i] = i;
}
int findF(int x){
    int z = x;
    while(x != father[x])x = father[x];
    while(z != father[z]){
        int temp = father[z];
        father[z] = x;
        z = temp;
    }
    return x;
}
void Union(int a, int b){
    int fa = findF(a), fb = findF(b);
    if(fa != fb)father[fb] = fa;
}
bool cmp(node a, node b){
    bool flag = false;
    if(a.Aav / a.peoNum > b.Aav / b.peoNum){
        flag = true;
    }else if(a.Aav / a.peoNum == b.Aav / b.peoNum && a.id < b.id){
        flag = true;
    }
    return flag;
}
int main(){
    init();
    int N = 0;
    vector<node>v, v2;
    scanf("%d", &N);
    for(int i = 0; i < N; ++i){
        int f, m, id, k, idm = 10000;
        int peonum = 0;
        scanf("%d%d%d%d", &id, &f, &m, &k);
        if(f == -1 && m == -1)idm = id;
        if(f == -1 && m != -1)idm = min(id, m);
        if(f != -1 && m == -1)idm = min(id, f);
        if(f != -1 && m != -1)idm = min(min(m, f), id);
        if(!vis[id]){
            peonum = 1;
            vis[id] = true;
        }
        if(f != -1){
            Union(f, id);
            if(!vis[f]){
                peonum++;
                vis[f] = true;
            }
        }
        if(m != -1){
            Union(m, id);
            if(!vis[m]){
                peonum++;
                vis[m] = true;
            }

        }
        for(int j = 0; j < k; ++j){
            int child;
            scanf("%d", &child);
            idm = min(idm, child);
            Union(id, child);
            if(!vis[child]){
                peonum++;
                vis[child] = true;
            }
        }
        int setNum, area;
        scanf("%d%d", &setNum, &area);
        int flag = false;
        for(int j = 0; j < v.size(); ++j){
            if(findF(v[j].id) == findF(idm)){
                v[j].id = min(idm, v[j].id);
                v[j].peoNum += peonum;
                v[j].Sav += (double)setNum;
                v[j].Aav += (double)area;
                flag = true;
                break;
            }
        }
        if(!flag)v.push_back({idm, peonum , (double)setNum, (double)area});
    }
    //录入数据的时候只合并了一部分家庭，再合并一遍
    v2.push_back(v[0]);
    for(int i = 1; i < v.size(); ++i){
        int flag = false;
        for(int j = 0; j < v2.size(); ++j){
            if(findF(v[i].id) == findF(v2[j].id)){
                v2[j].id = min(v[i].id, v2[j].id);
                v2[j].peoNum += v[i].peoNum;
                v2[j].Sav += v[i].Sav;
                v2[j].Aav += v[i].Aav;
                flag = true;
                break;
            }
        }
        if(!flag)v2.push_back(v[i]);
    }
    cout<<v2.size()<<endl;
    sort(v2.begin(), v2.end(), cmp);
    for(int i = 0; i < v2.size(); ++i){
        printf("%04d %d %.3f %.3f\n", v2[i].id, v2[i].peoNum, v2[i].Sav / v2[i].peoNum, v2[i].Aav / v2[i].peoNum);
    }
    return 0;
}