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 Nlines 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题目大意：找出一个家庭中人均有几套房产，和人均面积；输入中为 当前人的ID 父亲ID 母亲ID 当前人拥有几套 当前人拥有总面积。如果父母去世了，则用-1表示。输出要求按平均房产套数递减，如果相同，那么按ID递增排列。
//确实是使用并查集。
//遇到问题：遇到像8888这种单户的，该怎么处理，我写的里边似乎处理不了。
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include<cstdio>
using namespace std;
struct Peo{
int father,estate,area;
Peo(){father=-;}
}peo[];
map<int,int> mp;
struct Far{
int id,mem;
double avge,avga;
};
int findF(int a){
if(peo[a].father==-)return a;
int k=peo[a].father;
while(peo[k].father!=-){
peo[a].father=peo[k].father;
a=k;
k=peo[a].father;
//cout<<"wwww";
}
//cout<<"fff";
return k;
}
void unionF(int a,int b){
int fa=findF(a);
int fb=findF(b);
// if(fa>fb)peo[fa].father=fb;
// else peo[fb].father=fa;//这里出现了问题啊，得判断是否等于，等于的话，啥也不用了。
if(fa>fb)peo[fa].father=fb;
else if(fa<fb) peo[fb].father=fa;
//cout<<"uuuu";
} bool cmp(Far & a,Far & b){
return a.avge!=b.avge?a.id<b.id:a.avge>b.avge;
}
int main() {
int n;
cin>>n;
//fill(father,father+10000,-1);
int id,fa,mo,k,child;
for(int i=;i<n;i++){
cin>>id>>fa>>mo>>k;
//先将当前的人和父亲母亲合并
if(fa!=-)
unionF(id,fa);
if(mo!=-)
unionF(id,mo);
for(int j=;j<k;j++){
cin>>child;
unionF(id,child);
}
// peo[id].father=id;
cin>>peo[id].estate>>peo[id].area;
} vector<int> vt[];
for(int i=;i<;i++){
if(peo[i].father!=-){
//cout<<"hh";
mp[peo[i].father]++;
//怎么记录每个簇里有谁呢？
vt[peo[i].father].push_back(i);
}
} //最终还得sort一下。
cout<<mp.size()<<'\n';
vector<Far> far;
for(auto it=mp.begin();it!=mp.end();it++){
int id=it->first;
int mem=vt[id].size();
int tote,tota;
tote=peo[id].estate;
tota=peo[id].area;
for(int i=;i<vt[id].size();i++){
tote+=peo[vt[id][i]].estate;
tota+=peo[vt[id][i]].area;
}
far.push_back(Far{id,mem,tote*1.0/mem,tota*1.0/mem});
}
sort(far.begin(),far.end(),cmp);
for(int i=;i<far.size();i++){
printf("%04d %d %.3f %.3f\n",far[i].id,far[i].mem,far[i].avge,far[i].avga);
} return ;
}
//写成了这样，最终决定放弃！
代码转自：https://www.liuchuo.net/archives/2201
#include <cstdio>
#include <algorithm>
using namespace std;
struct DATA {
int id, fid, mid, num, area;
int cid[];//最多有10个孩子。
}data[];
struct node {
int id, people;
double num, area;
bool flag = false;
}ans[];
int father[];
bool visit[];
int find(int x) {
while(x != father[x])//这样真的好简便，我为啥写那么复杂呢。
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 = ;
scanf("%d", &n);
for(int i = ; i < ; i++)
father[i] = i;//将父亲设置为自己。
for(int i = ; i < n; i++) {
scanf("%d %d %d %d", &data[i].id, &data[i].fid, &data[i].mid, &k);
//直接读进来，不使用中间变量。
//并没有使用下标作为id啊。
visit[data[i].id] = true;//标记出现过了。
if(data[i].fid != -) {
visit[data[i].fid] = true;
Union(data[i].fid, data[i].id);//将id作为并查集中的关键字合并。
}
if(data[i].mid != -) {
visit[data[i].mid] = true;
Union(data[i].mid, data[i].id);
}
for(int j = ; 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 = ; i < n; i++) {
int id = find(data[i].id);//找到当前人的父亲，
ans[id].id = id;//现在这个ans中使用id作为下标索引了！
ans[id].num += data[i].num;
ans[id].area += data[i].area;
ans[id].flag = true;
}
for(int i = ; i < ; i++) {
if(visit[i])//如果它出现过。
i ans[find(i)].people++;
if(ans[i].flag)//标记有几簇人家。
cnt++;
}
for(int i = ; i < ; 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);
//并没有多个num属性，都是用一个存的，一开始就设为double。
//我居然还分开定义了。
}
}
sort(ans, ans + , cmp1);
printf("%d\n", cnt);
for(int i = ; i < cnt; i++)
printf("%04d %d %.3f %.3f\n", ans[i].id, ans[i].people, ans[i].num, ans[i].area);
return ;
}
1.将Father初始化为了自己，那么find函数就简单了，学习了
2.使用一个bool数组来标记出现过的点和没出现过的点，就解决了一家只有一口的情况。
3.在求ans向量的时候，仍使用了find，其实不用map的，find找到父亲都是可以用的。
4.因为data[i].id里存的是已经出现过的id,对那些没出现过的ans.flag肯定不会被标记为true的！
这是错误理解：ans里存储的所有的10000个人的情况，对于那些没有出现过的id，ans[i].flag也是true，只不过它所有的数据都是0，在经过排序之后，再通过簇进行控制输出，其他的不输出。
总之，学习了。