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 ... 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:

106666 5551 5552 1 7777 1 1001234 5678 9012 1 0002 2 3008888 -1 -1 0 1 10002468 0001 0004 1 2222 1 5007777 6666 -1 0 2 3003721 -1 -1 1 2333 2 1509012 -1 -1 3 1236 1235 1234 1 1001235 5678 9012 0 1 502222 1236 2468 2 6661 6662 1 3002333 -1 3721 3 6661 6662 6663 1 100

Sample Output:

38888 1 1.000 1000.0000001 15 0.600 100.0005551 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idint 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;}