标签：ext -- NPU print ima tput 没有 eal mem
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 0.750 100.000两个代码，仅供参考：
1 #include <iostream> 2 #include <vector> 3 #include <set> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 struct Node 8 { 9 int ID, nums = 0; 10 double ss = 0.0, aa = 0.0; 11 }; 12 int n; 13 double num[10000] = { 0.0 }, area[10000] = { 0.0 };//房子数量//房子面积 14 int mark[10000];//标记属于哪个家族 15 vector<set<int>>sets(10000);//记录每个家庭有多少人 16 vector<Node*>res; 17 void combin(int my, int other)//合并家族 18 { 19 for (auto ptr = sets[other].begin(); ptr != sets[other].end(); ++ptr)//将其他成员进行同化 20 { 21 mark[*ptr] = my; 22 sets[my].insert(*ptr); 23 } 24 sets[other].clear();//删除其他成员的标记 25 } 26 int main() 27 { 28 cin >> n; 29 fill(mark, mark + 10000, -1); 30 for (int i = 0; i < n; ++i) 31 { 32 int my, parent, k, child, nn, aa; 33 cin >> my; 34 if (mark[my] == -1)//还未标记是哪个家族 35 { 36 mark[my] = my;//就以自己为标记 37 sets[mark[my]].insert(my); 38 } 39 for (int i = 0; i < 2; ++i)//添加父母 40 { 41 cin >> parent; 42 if (parent == -1) 43 continue; 44 if (mark[parent] == -1)//家人未标记 45 { 46 mark[parent] = mark[my]; 47 sets[mark[my]].insert(parent); 48 } 49 else if (mark[parent] != -1 && mark[parent] != mark[my])//家人与自己标记不同 50 combin(mark[my], mark[parent]);//将家人同化为自己标记 51 } 52 cin >> k; 53 while (k--) 54 { 55 cin >> child; 56 if (mark[child] == -1)//孩子未标记 57 { 58 mark[child] = mark[my]; 59 sets[mark[my]].insert(child); 60 } 61 else if (mark[child] != -1 && mark[child] != mark[my])//孩子与自己标记不同 62 combin(mark[my], mark[child]);//将孩子同化为自己标记 63 } 64 cin >> nn >> aa; 65 num[my] = nn; 66 area[my] = aa; 67 } 68 for (int i = 0; i < sets.size(); ++i) 69 { 70 if (sets[i].empty()) 71 continue; 72 Node* temp = new Node; 73 temp->ID = *sets[i].begin(); 74 temp->nums = sets[i].size(); 75 for (auto ptr = sets[i].begin(); ptr != sets[i].end(); ++ptr) 76 { 77 temp->aa += area[*ptr]; 78 temp->ss += num[*ptr]; 79 } 80 temp->ss /= temp->nums; 81 temp->aa /= temp->nums; 82 res.push_back(temp); 83 } 84 sort(res.begin(), res.end(), [](Node*a, Node*b) { 85 if (abs(a->aa - b->aa) < 0.0001) 86 return a->ID < b->ID; 87 else 88 return a->aa > b->aa; }); 89 cout << res.size() << endl; 90 for (auto v : res) 91 printf("%04d %d %0.3f %0.3f\n", v->ID, v->nums, v->ss, v->aa); 92 return 0; 93 }
1 #include<bits/stdc++.h> 2 using namespace std; 3 struct Estate{//存储集合最小id、集合结点个数num、集合总sets、集合总area 4 int id,num=0; 5 double sets=0.0,area=0.0; 6 }; 7 bool cmp(const Estate&e1,const Estate&e2){//比较函数 8 return e1.area!=e2.area?e1.area>e2.area:e1.id<e2.id; 9 } 10 int father[10005];//并查集底层数组 11 int findFather(int x){//查找根节点 12 if(x==father[x]) 13 return x; 14 int temp=findFather(father[x]); 15 father[x]=temp; 16 return temp; 17 } 18 void unionSet(int a,int b){//合并两个集合 19 int ua=findFather(a),ub=findFather(b); 20 if(ua!=ub) 21 father[max(ua,ub)]=min(ua,ub);//以小的id作为根节点 22 } 23 unordered_map<int,pair<double,double>>idEstate; 24 unordered_map<int,Estate>familyEstate; 25 int main(){ 26 int N; 27 scanf("%d",&N); 28 iota(father,father+10005,0); 29 while(N--){//读取数据并合并相关集合 30 int id,f,m,k,a; 31 scanf("%d%d%d%d",&id,&f,&m,&k); 32 if(f!=-1) 33 unionSet(id,f); 34 if(m!=-1) 35 unionSet(id,m); 36 while(k--){ 37 scanf("%d",&a); 38 unionSet(id,a); 39 } 40 scanf("%lf%lf",&idEstate[id].first,&idEstate[id].second);//记录id和对应的sets、area 41 } 42 for(int i=0;i<10005;++i){//遍历并查集数组 43 int temp=findFather(i); 44 if(temp!=i)//根节点不等于本身 45 ++familyEstate[temp].num;//递增根节点下集合的结点个数 46 if(idEstate.find(i)!=idEstate.cend()){//累加集合的总sets、总area 47 familyEstate[temp].sets+=idEstate[i].first; 48 familyEstate[temp].area+=idEstate[i].second; 49 } 50 } 51 vector<Estate>v; 52 for(auto i=familyEstate.begin();i!=familyEstate.end();++i){//将familyEstate中的值搬迁到vector中,并更新相关信息 53 (i->second).id=i->first; 54 ++(i->second).num;//集合下结点个数没有统计根节点，所以要+1 55 (i->second).sets/=(i->second).num; 56 (i->second).area/=(i->second).num; 57 v.push_back(i->second); 58 } 59 sort(v.begin(),v.end(),cmp);//排序 60 printf("%d\n",v.size()); 61 for(int i=0;i<v.size();++i) 62 printf("%04d %d %.3f %.3f\n",v[i].id,v[i].num,v[i].sets,v[i].area); 63 return 0; 64 }
PAT甲级——A1114 Family Property【25】
标签：ext -- NPU print ima tput 没有 eal mem
原文地址：https://www.cnblogs.com/zzw1024/p/11456327.html
