PAT(A) - 1114. Family Property (25)

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 ... 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

思路分析：考查并查集，这道题很烦。

#include <cstdio>
#include <vector>
#include <algorithm>

#define MAX 10001
// 这里有个BUG，我把MAX写成10000的时候，9999的父节点就变成6663了
// 正常情况应该是9999，还不知道为啥

using namespace std;

typedef struct {
    int house;
    int area;
} Person;

typedef struct {        // 保存最后的结果数组，最后要进行排序
    int minPerson;      // 家族最小成员的编号
    int sumPerson;      // 家族所有人的数量
    int sumHouse;       // 家族一共有的房产数量
    int sumArea;        // 家族一共有的房产面积
    double aveHouse;    // 房产数量人均值
    double aveArea;     // 房产面积人均值
} Result;

Person p[MAX];          // 每个人的房产信息
int isRoot[MAX];        // 是否是根节点
int Father[MAX];        // 并查集数组
int visit[MAX];         // 结点是否访问过

void initPerson() {
    for( int i = 0; i < MAX; i++ ) {
        Father[i] = i;
        visit[i] = 0;
        isRoot[i] = 0;
        p[i].house = 0;
        p[i].area = 0;
    }
}

int findFather( int x ) {
    int a = x;
    while( x != Father[x] ) {
        x = Father[x];
    }

    // 路径压缩
    int z;
    while( a != Father[a] ) {
        z = a;
        a = Father[a];
        Father[z] = x;
    }

    return x;
}

void Union( int a, int b ) {
    int findA = findFather( a );
    int findB = findFather( b );

    if( findA != findB ) {
        Father[findA] = findB;
    }
    //printf( "%04d and %04d in set\n", a, b );
}

int cmp( Result a, Result b ) {
    if( a.aveArea != b.aveArea ) {
        return a.aveArea > b.aveArea ? 1 : 0;
    }
    return a.minPerson > b.minPerson ? 0 : 1;
}

int main() {
    //freopen( "123.txt", "r", stdin );
    int n;
    int id, father, mother, k, child, house, area;

    scanf( "%d", &n );
    initPerson();


    for( int i = 0; i < n; i++ ) {
        scanf( "%d%d%d%d", &id, &father, &mother, &k );
        visit[id] = visit[father] = visit[mother] = 1;
        if( father != -1 )
            Union( id, father );
        if( mother != -1 )
            Union( id, mother );

        while( k-- ) {
            scanf( "%d", &child );
            visit[child] = 1;
            Union( id, child );
        }
        scanf( "%d%d", &p[id].house, &p[id].area );
    }

    for( int i = 0; i < MAX; i++ ) {
        if( visit[i] ) {
            int fa = findFather( i );
            //printf( "%04d's father = %04d\n", i, fa );
            isRoot[fa] = 1;
        }
    }

    //printf( "\n" );
    vector<int> vec;
    for( int i = 0; i < MAX; i++ ) {
        if( isRoot[i] ) {
            //printf( "%d\n", i );
            vec.push_back( i );
        }
    }

    vector<Result> vecR;
    for( int i = 0; i < vec.size(); i++ ) {
        int fa = vec[i];
        bool flag = false;
        int min;
        int sumPerson = 0;
        int sumArea = 0;
        int sumHouse = 0;
        for( int i = 0; i < MAX; i++ ) {
            if( fa == findFather( i ) ) {
                if( !flag ) {
                    min = i;
                    flag = true;
                }
                sumPerson++;
                sumHouse += p[i].house;
                sumArea += p[i].area;
            }
        }
        Result r;
        r.minPerson = min;
        r.sumPerson = sumPerson;
        r.sumHouse = sumHouse;
        r.sumArea = sumArea;
        r.aveHouse = (double)sumHouse / (double)sumPerson;
        r.aveArea = (double)sumArea / (double)sumPerson;
        vecR.push_back( r );
    }

    sort( vecR.begin(), vecR.end(), cmp );

    printf( "%d\n", vecR.size() );
    for( int i = 0; i < vecR.size(); i++ ) {
        printf( "%04d %d %.3lf %.3lf\n",
            vecR[i].minPerson, vecR[i].sumPerson, vecR[i].aveHouse, vecR[i].aveArea );
    }
    return 0;
}