PAT 1114 Family Property
作者：互联网
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 (≤). 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) 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
思路为 使用并查集，确定每个家族中的最小ID，因为只有输入数据具有area set等信息，第一次遍历将房产信息归到 家族中最小ID的名下，顺便标记为是最小ID；
第二次遍历，将出现过的ID找到，给其对应的家族 的人数 加一； 同时记录有多少最小ID，即有多少个家族；
第三次遍历，将平均值设置好，压入vector中，最后排序
using namespace std;
int pre[10000];
struct node{
int id,s,a;
};
int find(int x){
return pre[x]==x?x:find(pre[x]);
}
void uni(int a,int b){
int faa=find(a);
int fab=find(b);
if(faa>fab){
pre[faa]=fab;
}else{
pre[fab]=faa;
}
}
struct eva{
int head,rs;
double s,a,avs,ava;
bool tou;
};
eva final[10004];
bool cmp1(eva a,eva b){
if(a.ava!=b.ava){
return a.ava>b.ava;
}else{
return a.head<b.head;
}
}
bool visit[10000];
vector<node> arr;
int main() {
//freopen("D://dio.txt","r",stdin);
int n,id,fa,ma,k,kid,sets,area,i,j;
for(i=0;i<10000;i++){
pre[i]=i;
}
cin>>n;
int fn,ff,fm,fk;
node temp;
for(i=0;i<n;i++){
scanf("%d %d %d %d",&id,&fa,&ma,&k);
fn=find(id);
if(fa!=-1){
ff=find(fa);
uni(fa,id);
visit[fa]=true;
}
if(ma!=-1){
fm=find(ma);
uni(ma,id);
visit[ma]=true;
}
visit[id]=true;
for(j=0;j<k;j++){
scanf("%d",&kid);
fk=find(kid);
uni(fk,fn);
visit[kid]=true;
}
scanf("%d %d",&sets,&area);
temp.id=id;
temp.s=sets;
temp.a=area;
arr.push_back(temp);
}
for(i=0;i<arr.size();i++){
int ii=find(arr[i].id);
final[ii].head=ii;
final[ii].s+=arr[i].s;
final[ii].a+=arr[i].a;
final[ii].tou=true;
}
int cnt=0;
vector<eva> ginto;
for(i=0;i<10000;i++){
if(visit[i]){
final[find(i)].rs++;
}
if(final[i].tou){
cnt++;
}
}
for(i=0;i<10000;i++){
if(final[i].tou){
final[i].avs=final[i].s/final[i].rs;
final[i].ava=final[i].a/final[i].rs;
ginto.push_back(final[i]);
}
}
cout<<cnt<<endl;
sort(ginto.begin(),ginto.end(),cmp1);
for(i=0;i<ginto.size();i++){
printf("%04d %d %.3f %.3f\n",ginto[i].head,ginto[i].rs,ginto[i].avs,ginto[i].ava);
}
return 0;
}
标签：PAT,family,estate,area,int,number,1114,Property,ID 来源： https://www.cnblogs.com/xfdmyghdx/p/10366802.html