PAT 甲级 A1114 Family Property(并查集)

PAT Advanced Level 1114 Family Property

今天磕了蛮久才把这道题磕出来。。

这道题用并查集做就可以。传统的并查集是int father[]数组，这里我们找最小值。

所以合并的Union函数里不是直接father[fa]=fb，而是用father[max(fa,fb)]=min(fa,fb)。当然我们这里数组名不用father表示，我采用了minID作为数组名。

我的minID都初始化为-1，每次出现新的ID时，更新它的minID值为这个ID。最后遍历数组，不是-1的都是出现过的结点

 

需要注意的点和学到的东西：

①sort函数可以不用自己写比较函数，重载operator<即可。

②double类型不能用!=判断两个值是否相等，但用<和>是可以的。

③node结点里无需维护父子结点，如果在我的代码里加上这两样关系，会报内存超限的错误……

 

题目描述：

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​1​​⋯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

代码如下：

#include 
#include 
#include 
#include 
using namespace std;
const int maxn=10005;
int minID[maxn];    //存储每个node的最小ID 是并查集的集

struct Node{//每个node表示一个人
    int M=0,area=0;
    int ID;
}node[maxn];

struct family{
    int ID;    //表示家庭中最小的ID
    int cnt=1;    //表示家庭的人数
    int area=0;    //表示家庭总面积
    int M=0;    //表示家庭总房产个数
    bool operator<(const family tmp)const{ //重载比较符号后可直接使用sort默认排序函数
        double a=double(area)/cnt;
        double b=double(tmp.area)/tmp.cnt;
        if(a>b||ab;
        return ID F;
    for(int i=0;i