1114. Family Property (25)-PAT甲级真题(并查集)

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


题目理解

  • 给出每个家庭的信息和房产信息
    • 本人ID
    • 父亲ID
    • 母亲ID
    • 小孩个数
      • 小孩id
    • 拥有房产数
    • 拥有房产面积
  • 计算整个家族的平均房产信息
    • 家庭最小成员编号
    • 家庭成员个数
    • 总房产数
    • 平均房产数
    • 平均面积
  • 家庭信息按照平均面积进行降序排序,平均面积相同时按照序号升序输出,平均值保留3位小数

分析

数据结构和函数

  • 设置一个DATA结构体存储输入信息,设置ans结构体存储输出结果
  • 设置UNION函数将同一个家族的成员进行合并,find函数用于查找root节点
  • 设置比较规则cmp1用于结果排序,平均面积进行降序排序,平均面积相同时按照序号升序输出
  • visit[10000]数组用于记录哪些id出现过

逻辑

  • 接收输入,并将id与mid,fid,cid[]进行union操作
  • 统计各家族的房产数量和房产面积
  • 统计各家族的人数(累加所有visit[i]==1,ans[find(i)].people++),统计家族个数(ans[i].flag==1, cnt++)
  • 统计平均房产数和面积
  • 将结果进行排序,使用cmp1规则
  • 结果输出,平均数保留3位小数
#include 
#include 

using namespace std;
struct DATA {
    int id, fid, mid, num, area;
    int cid[10];
} data[1001];
struct node {
    int id, people;
    double num, area;
    bool flag = false;  //root节点flag=true
} ans[10000];

int father[10000];
int visit[10000];

int find(int x) {
    while (x != father[x]) {
        x = father[x];
    }
    return x;
}

void Union(int a, int b) {
    int faA = find(a);
    int faB = find(b);
    if (faA > faB)
        father[faA] = faB;
    else
        father[faB] = faA;
}

//多规则排序
int cmp1(node a, node b) {
    if (a.area != b.area)
        return a.area > b.area;
    else
        return a.id < b.id;
}

int main() {
    int n, k, cnt = 0;
    cin >> n;
    for (int i = 0; i < 10000; i++) {
        father[i] = i;
    }
    //接收输入,并进行Union操作
    for (int i = 0; i < n; i++) {
        cin >> data[i].id >> data[i].fid >> data[i].mid >> k;
        visit[data[i].id] = true;
        if (data[i].fid != -1) {
            visit[data[i].fid] = true;
            Union(data[i].id, data[i].fid);
        }
        if (data[i].mid != -1) {
            visit[data[i].mid] = true;
            Union(data[i].id, data[i].mid);
        }
        for (int j = 0; j < k; j++) {
            cin >> data[i].cid[j];
            visit[data[i].cid[j]] = true;
            Union(data[i].id, data[i].cid[j]);
        }
        cin >> data[i].num >> data[i].area;
    }
    //累计各家族的房产数量,面积
    for (int i = 0; i < n; i++) {
        int root = find(data[i].id);
        ans[root].id = root;
        ans[root].num += data[i].num; //累加家族房产数量
        ans[root].area += data[i].area;  //累加家族面积
        ans[root].flag = true;

    }

    for (int i = 0; i < 10000; i++) {
        if (visit[i])
            ans[find(i)].people++; //各家族人数统计
        if (ans[i].flag)
            cnt++;  //家族个数统计
    }
    //统计平均房产数和面积
    for (int i = 0; i < 10000; i++) {
        if (ans[i].flag) {
            ans[i].num = (double) (ans[i].num / ans[i].people);
            ans[i].area = (double) (ans[i].area / ans[i].people);
        }
    }
    sort(ans, ans + 10000, cmp1);
    cout << cnt << endl;
    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);
    }

}

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