1114. Family Property (25) <并查集,set>

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

一行纪录所有节点的父亲都是find(id) 构造数组
房产和面积绑定给id

将所有纪录放在set里面，遍历，找到所有的父亲节点。

并且计算总的房产和面积

最后在遍历找到最小的id，排序输出

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int k;
int father[10000];
int sum[10000]={0},area[10000]={0};
int fa[10000]={0};
typedef struct node{
int id,sum;
double a_set,a_area;
}node;
int find(int x){
if(x==father[x]) return x;
return father[x]=find(father[x]);
}
bool cmp(node n1,node n2){
if(n1.a_area!=n2.a_area){
return n1.a_area>n2.a_area;
}
return n1.id<n2.id;
}
int main(){
cin>>k;
set<int> nu;
for(int i=0;i<10000;i++) father[i]=i;
for(int i=0;i<k;i++){
int id,f,m;
scanf("%d%d%d",&id,&f,&m);
int fz=id;
nu.insert(id);
id=find(id);
if(f!=-1) {
father[find(f)]=id;
nu.insert(f);
}

if(m!=-1){
father[find(m)]=id;
nu.insert(m);
}
int num;
scanf("%d",&num);
for(int i=0;i<num;i++){
int chi;
scanf("%d",&chi);
nu.insert(chi);
father[find(chi)]=id;
}
int M_estate,Area;
scanf("%d%d",&M_estate,&Area);
sum[fz]=M_estate;
area[fz]=Area;
}
set<int> s;
for(set<int>::iterator it=nu.begin();it!=nu.end();it++){
int num=find(*it);
if(*it!=num) {
area[num]+=area[*it];
sum[num]+=sum[*it];
}

fa[num]++;
s.insert(num);
}
node no[1001];
int cnt=0;
int v[10000]={0};
for(set<int>::iterator it=nu.begin();it!=nu.end();it++){
int fz=find(*it);
if(v[fz]==0&&s.find(fz)!=s.end()){
v[fz]=1;
no[cnt].id=*it;
no[cnt].sum=fa[fz];
no[cnt].a_set=1.0*sum[fz]/fa[fz];
no[cnt].a_area=1.0*area[fz]/fa[fz];
cnt++;
}
}
/*for(int i=0;i<=9999;i++){ //这点是想到set容器默认从小到大排序，这样会浪费时间，直接遍历set容器寻找最小id
int fz=find(i);
if(v[fz]==0&&s.find(fz)!=s.end()){
v[fz]=1;
no[cnt].id=i;
no[cnt].sum=fa[fz];
no[cnt].a_set=1.0*sum[fz]/fa[fz];
no[cnt].a_area=1.0*area[fz]/fa[fz];
cnt++;
//printf("%04d %d %.3f %.3f\n",i,fa[fz],1.0*sum[fz]/fa[fz],1.0*area[fz]/fa[fz]);
}
} */
sort(no,no+cnt,cmp);
cout<<cnt<<endl;
for(int i=0;i<cnt;i++){
printf("%04d %d %.3f %.3f\n",no[i].id,no[i].sum,no[i].a_set,no[i].a_area);
}
return 0;
}