1114 Family Property (25分) （并查集）复杂题经典并查集

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 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

#include <iostream>
#include <vector>
#include <algorithm>using namespace std;struct Data
{int id, fid, mid, num, area;int cid[10];
};struct person
{int id, people;double num, area;bool flag = false;
}ans[10000];bool cmp(person a, person b)
{if (a.area!=b.area){return a.area > b.area;}else{return a.id < b.id;}
}vector<int> father(10000);
vector<bool> visit(10000);int find(int x)
{if (x==father[x])//含路径压缩{return x;}return father[x] = find(father[x]);/*while (x!=father[x]){x = father[x];}return  x;*/
}void Union(int a,int b)
{int fA = find(a), fB = find(b);if (fA>fB){father[fA] = fB;}else if (fA<fB){father[fB] = fA;}
}int main()
{//freopen("in.txt", "r", stdin);int n, k, cnt = 0;cin >> n;for (int i = 0; i < 10000; i++)//初始化{father[i] = i;}vector<Data> d(1000);for (int i = 0; i < n; i++){cin >> d[i].id >> d[i].fid >> d[i].mid >> k;visit[d[i].id] = true;if (d[i].fid!=-1){visit[d[i].fid] = true;Union(d[i].id, d[i].fid);}if (d[i].mid!=-1){visit[d[i].mid] = true;Union(d[i].id, d[i].mid);}for (int j = 0; j < k; j++){cin >> d[i].cid[j];visit[d[i].cid[j]] = true;Union(d[i].id, d[i].cid[j]);          }cin >> d[i].num >> d[i].area;}for (int i = 0; i < n; i++){int id = find(d[i].id);ans[id].id = id;ans[id].num += d[i].num;ans[id].area += d[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, cmp);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);}return 0;
}

二刷：

#include <iostream>
#include <vector>
#include <algorithm>using namespace std;const int maxn = 10006;
vector<int> father(maxn);int findfather(int x)
{return x == father[x] ? x : (father[x] = findfather(father[x]));
}void Union(int a, int b)
{int fa = findfather(a), fb = findfather(b);if (fa>fb){father[fa] = fb;}else{father[fb] = fa;}
}struct data
{int id, fid, mid, estate, area;int cid[10];
}d[maxn];struct node
{int id,  num;double estate, area;bool flag;
}ans[maxn];bool cmp(node a, node b)
{if (a.area!=b.area){return a.area > b.area;}else{return a.id < b.id;}
}int main()
{//freopen("in.txt", "r", stdin);int n;cin >> n;for (int i = 0; i < maxn; i++){father[i] = i;}bool visit[maxn] = { false };for (int i = 1; i <= n; i++){cin >> d[i].id >> d[i].fid >> d[i].mid;visit[d[i].id] = true;if (d[i].fid != -1){visit[d[i].fid] = true;Union(d[i].id, d[i].fid);}if (d[i].mid != -1){visit[d[i].mid] = true;Union(d[i].id, d[i].mid);}int k;cin >> k;for (int j = 0; j < k; j++){cin >> d[i].cid[j];visit[d[i].cid[j]] = true;Union(d[i].id, d[i].cid[j]);}cin >> d[i].estate >> d[i].area;}for (int i = 1; i <= n; i++){int root = findfather(d[i].id);ans[root].id = root;ans[root].estate += d[i].estate;ans[root].area += d[i].area;ans[root].flag = true;}int cnt = 0;//family 个数for (int i = 0; i < maxn; i++){if (visit[i]){int root = findfather(i);ans[root].num++;}if (ans[i].flag){cnt++;}}for (int i = 0; i < maxn; i++){if (ans[i].flag){ans[i].estate = (double)(ans[i].estate / ans[i].num);ans[i].area = (double)(ans[i].area / ans[i].num);}}sort(ans, ans+maxn, cmp);cout << cnt << endl;for (int i = 0; i < cnt; i++){printf("%04d %d %.3f %.3f\n", ans[i].id, ans[i].num, ans[i].estate, ans[i].area);}return 0;
}