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（房產）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 Child
1
⋯Child
k
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; Child
i
'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
並查集模板
father[]
type find(type x)
{
while(x!=father[x])x=father[x];
return x;
}
void Union(type a,type b)
{
type fa=find(a);
type fb=find(b);
father[fa]=fb;////////////無所謂大小
/*小序號作爲老父親標記
if(fa>fb)
{
father[fa]=fb;
}
else
{
father[fb]=fa;
}
*/
}
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,id,fa,mo,k,t;
int father[10005];
bool vis[10005];
int no[10005];
struct Owner
{
int ID;
int sets;
int area;
}O[10005];
struct Family
{
int ID;
int num;
double sets;
double area;
}ans[10005];
int find(int x)
{
while(x!=father[x])x=father[x];
return x;
}
void Union(int a,int b)
{
int fa=find(a);
int fb=find(b);
if(fa<fb)
{
father[fb]=fa;
}
else
{
father[fa]=fb;
}
}
bool cmp(Family a,Family b)
{
if(a.area==b.area)
return a.ID<b.ID;
else
return a.area>b.area;
}
int main()
{
for(int i=0;i<10000;i++)father[i]=i;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d%d%d%d",&id,&fa,&mo,&k);
O[id].ID=id;
vis[id]=1;
if(fa!=-1)
{
vis[fa]=1;
Union(id,fa);
}
if(mo!=-1)
{
vis[mo]=1;
Union(id,mo);
}
for(int j=0;j<k;j++)
{
scanf("%d",&t);
vis[t]=1;
Union(id,t);
}
scanf("%d%d",&O[id].sets,&O[id].area);
}
int fami=0;
for(int i=0;i<10000;i++)
{
if(!vis[i])continue;
t=find(i);
if(!no[t])
{
no[t]=++fami;
ans[no[t]].ID=t;
}
ans[no[t]].num++;
ans[no[t]].sets+=O[i].sets;
ans[no[t]].area+=O[i].area;
}
for(int i=1;i<=fami;i++)/////////////////不能等輸出時再除，因爲排序要用到/num的結果值；
{
ans[i].sets/=ans[i].num;
ans[i].area/=ans[i].num;
}
sort(ans+1,ans+1+fami,cmp);
printf("%d\n",fami);
for(int i=1;i<=fami;i++)
{
printf("%04d %d %.3f %.3f\n",ans[i].ID,ans[i].num,ans[i].sets,ans[i].area);
}
return 0;
}