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

题目大意：给定每个人的家庭成员和其自己名下的房产，请你统计出每个家庭的人口数、人均房产面积及房产套数。首先在第一行输出家庭个数（所有有亲属关系的人都属于同一个家庭）。随后按下列格式输出每个家庭的信息：家庭成员的最小编号家庭人口数人均房产套数人均房产面积。其中人均值要求保留小数点后3位。家庭信息首先按人均面积降序输出，若有并列，则按成员编号的升序输出。
分析：用并查集。分别用两个结构体数组，一个data用来接收数据，接收的时候顺便实现了并查集的操作union，另一个数组ans用来输出最后的答案，因为要计算家庭人数，所以用visit标记所有出现过的结点，对于每个结点的父结点，people++统计人数。标记flag == true，计算true的个数cnt就可以知道一共有多少个家庭。排序后输出前cnt个就是所求答案～～

#include <cstdio>
#include <algorithm>
using namespace std;
struct DATA {
    int id, fid, mid, num, area;
    int cid[10];
}data[1005];
struct node {
    int id, people;
    double num, area;
    bool flag = false;
}ans[10000];
int father[10000];
bool 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 if(faA < faB)
        father[faB] = faA;
}
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;
    scanf("%d", &n);
    for(int i = 0; i < 10000; i++)
        father[i] = i;
    for(int i = 0; i < n; i++) {
        scanf("%d %d %d %d", &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].fid, data[i].id);
        }
        if(data[i].mid != -1) {
            visit[data[i].mid] = true;
            Union(data[i].mid, data[i].id);
        }
        for(int j = 0; j < k; j++) {
            scanf("%d", &data[i].cid[j]);
            visit[data[i].cid[j]] = true;
            Union(data[i].cid[j], data[i].id);
        }
        scanf("%d %d", &data[i].num, &data[i].area);
    }
    for(int i = 0; i < n; i++) {
        int id = find(data[i].id);
        ans[id].id = id;
        ans[id].num += data[i].num;
        ans[id].area += data[i].area;
        ans[id].flag = true;
    }
    for(int i = 0; i < 10000; i++) {
        if(visit[i])
            ans[find(i)].people++;
        if(ans[i].flag)
            cnt++;
    }
    for(int i = 0; i < 10000; i++) {
        if(ans[i].flag) {
            ans[i].num = (double)(ans[i].num * 1.0 / ans[i].people);
            ans[i].area = (double)(ans[i].area * 1.0 / ans[i].people);
        }
    }
    sort(ans, ans + 10000, cmp1);
    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 <algorithm>

using namespace std;

struct DATA {

int id, fid, mid, num, area;

int cid[10];

}data[1005];

struct node {

int id, people;

double num, area;

bool flag = false;

}ans[10000];

int father[10000];

bool 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 if(faA < faB)

father[faB] = faA;

}

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;

scanf("%d", &n);

for(int i = 0; i < 10000; i++)

father[i] = i;

for(int i = 0; i < n; i++) {

scanf("%d %d %d %d", &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].fid, data[i].id);

}

if(data[i].mid != -1) {

visit[data[i].mid] = true;

Union(data[i].mid, data[i].id);

}

for(int j = 0; j < k; j++) {

scanf("%d", &data[i].cid[j]);

visit[data[i].cid[j]] = true;

Union(data[i].cid[j], data[i].id);

}

scanf("%d %d", &data[i].num, &data[i].area);

}

for(int i = 0; i < n; i++) {

int id = find(data[i].id);

ans[id].id = id;

ans[id].num += data[i].num;

ans[id].area += data[i].area;

ans[id].flag = true;

}

for(int i = 0; i < 10000; i++) {

if(visit[i])

ans[find(i)].people++;

if(ans[i].flag)

cnt++;

}

for(int i = 0; i < 10000; i++) {

if(ans[i].flag) {

ans[i].num = (double)(ans[i].num * 1.0 / ans[i].people);

ans[i].area = (double)(ans[i].area * 1.0 / ans[i].people);

}

sort(ans, ans + 10000, cmp1);

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;

}

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