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.
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 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) 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.
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.
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
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; }
原文：https://www.cnblogs.com/xfdmyghdx/p/10366802.html