1114 Family Property (PAT甲级)

#include 
#include 
#include 
#include 
#include 
const int MAXN = 10000;

struct node{
    int id;
    int memberNbr = 1;
    int setNbr;
    int area;
    node(int _id, int _setNbr, int _area): id(_id), setNbr(_setNbr), area(_area){}
};

int N, id, father, mother, k, t, M, area, cnt;
int fa[MAXN];
std::map mpSetNbr, mpArea, mpCnt;
std::set st;
std::vector vec;

int findFather(int x){
    int k = x;
    while(fa[x] != x){
        x = fa[x];
    }
    int t;
    while(fa[k] != k){
        t = fa[k];
        fa[k] = x;
        k = t;
    }
    return x;
}

void Union(int a, int b){
    int faA = findFather(a);
    int faB = findFather(b);
    if(faA < faB){
        fa[faB] = faA;
    } else{
        fa[faA] = faB;
    }
}

bool cmp(const node &a, const node &b){
    double avgA = a.area * 1.0 / a.memberNbr;
    double avgB = b.area * 1.0 / b.memberNbr;
    return avgA != avgB ? avgA > avgB : a.id < b.id;
}

int main(){
    for(int i = 0; i < MAXN; ++i){
        fa[i] = i;
    }
    scanf("%d", &N);
    for(int i = 0; i < N; ++i){
        scanf("%d %d %d %d", &id, &father, &mother, &k);
        st.insert(id);
        if(father != -1){
            Union(id, father);
            st.insert(father);
        }
        if(mother != -1){
            Union(id, mother);
            st.insert(mother);
        }
        for(int j = 0; j < k; ++j){
            scanf("%d", &t);
            Union(id, t);
            st.insert(t);
        }
        scanf("%d %d", &M, &area);
        mpSetNbr[id] = M;
        mpArea[id] = area;
    }
    cnt = 0;
    for(auto it = st.begin(); it != st.end(); ++it){
        if(*it == findFather(*it)){
            vec.push_back(node(*it, mpSetNbr[*it], mpArea[*it]));
            mpCnt.insert({*it, cnt++});
        } else{
            int i = mpCnt[findFather(*it)];
            vec[i].memberNbr++;
            vec[i].setNbr += mpSetNbr[*it];
            vec[i].area += mpArea[*it];
        }
    }
    sort(vec.begin(), vec.end(), cmp);
    printf("%d\n", vec.size());
    for(int i = 0; i < vec.size(); ++i){
        printf("%04d %d %.3f %.3f\n", vec[i].id, vec[i].memberNbr,
        vec[i].setNbr * 1.0 / vec[i].memberNbr, vec[i].area * 1.0 / vec[i].memberNbr);
    }
    return 0;
}

题目如下:

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​ Mestate​ 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; 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

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