1114 Family Property （25 分）
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 child2 ... childk M 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; 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:
\(ID M AVG_{sets} AVG_{area}\)
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
分析：要将每个家庭的数据分离出来，使用并查集即可实现，数据处理上有两种方法：
1.先将录入的数据按照输入的形式存在数组中，录入的同时完成集合的合并，然后再从录入的数据中挑出需要的数据，最后对挑出的数据仓晒礼输出；
2.在录入的同时进行集合的合并，并将数据按照输出的格式存储，然后进行处理
#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<unordered_map>
#include<set>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;
const int nmax = 10100;
int fath[nmax];
void init(){
for(int i = 0; i < nmax; ++i)fath[i] = i;
}
int findF(int x){
int z = x;
while(x != fath[x])x = fath[x];
while(z != fath[z]){
int temp = fath[z];
fath[z] = x;
z = temp;
}
return x;
}
void Union(int a, int b){
int fa = findF(a), fb = findF(b);
if(fa < fb)fath[fb] = fa;
else if(fa > fb)fath[fa] = fb;
}
struct node{
int id, people;
double num, area;
bool flag = false;
bool operator < (node &a)const{
return area != a.area ? area > a.area : id < a.id;
}
}ans[nmax];
struct data{
int id, fid, mid, k;
int child[6];
int num, area;
}v[nmax];
bool vis[nmax] = {false};
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("input.txt", "r", stdin);
#endif // ONLINE_JUDGE
init();
int N;
scanf("%d", &N);
//录入数据并合并，由于序号不连续，要标记哪些序号已经出现过
for(int i = 0; i < N; ++i){
int id, fid, mid, k;
scanf("%d%d%d%d", &id, &fid, &mid, &k);
v[id].id = id;
v[id].fid = fid;
v[id].mid = mid;
v[id].k = k;
vis[id] = true;
if(fid != -1){
Union(id, fid);
vis[fid] = true;
}
if(mid != -1){
Union(id, mid);
vis[mid] = true;
}
for(int j = 0; j < k; ++j){
scanf("%d", &v[id].child[j]);
Union(id, v[id].child[j]);
vis[v[id].child[j]] = true;
}
scanf("%d%d", &v[id].num, &v[id].area);
}
//统计并抽出需要的数据，用flag标记该根节点是否已经出现
int cnt = 0;
for(int i = 0; i < nmax; ++i){
if(vis[i] == true){
int index = findF(i);
ans[index].id = index;
ans[index].people++;
ans[index].num += v[i].num;
ans[index].area += v[i].area;
if(ans[index].flag == false)cnt++;
ans[index].flag = true;
}
}
//处理
for(int i = 0; i < nmax; ++i){
if(ans[i].flag == true){
ans[i].num /= ans[i].people;
ans[i].area /= ans[i].people;
}
}
//排序输出
sort(ans, ans + nmax);
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<iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
const int Nmax = 10000;
int father[Nmax];
bool vis[Nmax] = {false};
struct node{
int id, peoNum;
double Sav, Aav;
};
void init(){
for(int i = 0; i < Nmax; ++i)father[i] = i;
}
int findF(int x){
int z = x;
while(x != father[x])x = father[x];
while(z != father[z]){
int temp = father[z];
father[z] = x;
z = temp;
}
return x;
}
void Union(int a, int b){
int fa = findF(a), fb = findF(b);
if(fa != fb)father[fb] = fa;
}
bool cmp(node a, node b){
bool flag = false;
if(a.Aav / a.peoNum > b.Aav / b.peoNum){
flag = true;
}else if(a.Aav / a.peoNum == b.Aav / b.peoNum && a.id < b.id){
flag = true;
}
return flag;
}
int main(){
init();
int N = 0;
vector<node>v, v2;
scanf("%d", &N);
for(int i = 0; i < N; ++i){
int f, m, id, k, idm = 10000;
int peonum = 0;
scanf("%d%d%d%d", &id, &f, &m, &k);
if(f == -1 && m == -1)idm = id;
if(f == -1 && m != -1)idm = min(id, m);
if(f != -1 && m == -1)idm = min(id, f);
if(f != -1 && m != -1)idm = min(min(m, f), id);
if(!vis[id]){
peonum = 1;
vis[id] = true;
}
if(f != -1){
Union(f, id);
if(!vis[f]){
peonum++;
vis[f] = true;
}
}
if(m != -1){
Union(m, id);
if(!vis[m]){
peonum++;
vis[m] = true;
}
}
for(int j = 0; j < k; ++j){
int child;
scanf("%d", &child);
idm = min(idm, child);
Union(id, child);
if(!vis[child]){
peonum++;
vis[child] = true;
}
}
int setNum, area;
scanf("%d%d", &setNum, &area);
int flag = false;
for(int j = 0; j < v.size(); ++j){
if(findF(v[j].id) == findF(idm)){
v[j].id = min(idm, v[j].id);
v[j].peoNum += peonum;
v[j].Sav += (double)setNum;
v[j].Aav += (double)area;
flag = true;
break;
}
}
if(!flag)v.push_back({idm, peonum , (double)setNum, (double)area});
}
//录入数据的时候只合并了一部分家庭，再合并一遍
v2.push_back(v[0]);
for(int i = 1; i < v.size(); ++i){
int flag = false;
for(int j = 0; j < v2.size(); ++j){
if(findF(v[i].id) == findF(v2[j].id)){
v2[j].id = min(v[i].id, v2[j].id);
v2[j].peoNum += v[i].peoNum;
v2[j].Sav += v[i].Sav;
v2[j].Aav += v[i].Aav;
flag = true;
break;
}
}
if(!flag)v2.push_back(v[i]);
}
cout<<v2.size()<<endl;
sort(v2.begin(), v2.end(), cmp);
for(int i = 0; i < v2.size(); ++i){
printf("%04d %d %.3f %.3f\n", v2[i].id, v2[i].peoNum, v2[i].Sav / v2[i].peoNum, v2[i].Aav / v2[i].peoNum);
}
return 0;
}