PAT 1114 Family Property[並查集][難]
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 Nlines 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
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題目大意：找出一個家庭中人均有幾套房產，和人均面積；輸入中為 當前人的ID 父親ID 母親ID 當前人擁有幾套 當前人擁有總面積。如果父母去世了，則用-1表示。輸出要求按平均房產套數遞減，如果相同，那麽按ID遞增排列。
//確實是使用並查集。
//遇到問題：遇到像8888這種單戶的，該怎麽處理，我寫的裏邊似乎處理不了。
#include <iostream> #include <algorithm> #include <vector> #include <map> #include<cstdio> using namespace std; struct Peo{ int father,estate,area; Peo(){father=-1;} }peo[10000]; map<int,int> mp; struct Far{ int id,mem; double avge,avga; }; int findF(int a){ if(peo[a].father==-1)return a; int k=peo[a].father; while(peo[k].father!=-1){ peo[a].father=peo[k].father; a=k; k=peo[a].father; //cout<<"wwww"; } //cout<<"fff"; return k; } void unionF(int a,int b){ int fa=findF(a); int fb=findF(b); // if(fa>fb)peo[fa].father=fb; // else peo[fb].father=fa;//這裏出現了問題啊，得判斷是否等於，等於的話，啥也不用了。 if(fa>fb)peo[fa].father=fb; else if(fa<fb) peo[fb].father=fa; //cout<<"uuuu"; } bool cmp(Far & a,Far & b){ return a.avge!=b.avge?a.id<b.id:a.avge>b.avge; } int main() { int n; cin>>n; //fill(father,father+10000,-1); int id,fa,mo,k,child; for(int i=0;i<n;i++){ cin>>id>>fa>>mo>>k; //先將當前的人和父親母親合並 if(fa!=-1) unionF(id,fa); if(mo!=-1) unionF(id,mo); for(int j=0;j<k;j++){ cin>>child; unionF(id,child); } // peo[id].father=id; cin>>peo[id].estate>>peo[id].area; } vector<int> vt[10000]; for(int i=0;i<10000;i++){ if(peo[i].father!=-1){ //cout<<"hh"; mp[peo[i].father]++; //怎麽記錄每個簇裏有誰呢？ vt[peo[i].father].push_back(i); } } //最終還得sort一下。 cout<<mp.size()<<‘\n‘; vector<Far> far; for(auto it=mp.begin();it!=mp.end();it++){ int id=it->first; int mem=vt[id].size(); int tote,tota; tote=peo[id].estate; tota=peo[id].area; for(int i=0;i<vt[id].size();i++){ tote+=peo[vt[id][i]].estate; tota+=peo[vt[id][i]].area; } far.push_back(Far{id,mem,tote*1.0/mem,tota*1.0/mem}); } sort(far.begin(),far.end(),cmp); for(int i=0;i<far.size();i++){ printf("%04d %d %.3f %.3f\n",far[i].id,far[i].mem,far[i].avge,far[i].avga); } return 0; }View Code
//寫成了這樣，最終決定放棄！
代碼轉自：https://www.liuchuo.net/archives/2201
#include <cstdio> #include <algorithm> using namespace std; struct DATA { int id, fid, mid, num, area; int cid[10];//最多有10個孩子。 }data[1005]; struct node { int id, people; double num, area; bool flag = false; }ans[10000]; int father[10000]; bool visit[10000]; int find(int x) { while(x != father[x])//這樣真的好簡便，我為啥寫那麽復雜呢。 x = father[x]; return x; } void Union(int a, int b) { int faA = find(a); int faB = find(b); if(faA > faB) father[faA] = faB; else if(faA < faB) father[faB] = faA; } int cmp1(node a, node b) { if(a.area != b.area) return a.area > b.area; else return a.id < b.id; } int main() { int n, k, cnt = 0; scanf("%d", &n); for(int i = 0; i < 10000; i++) father[i] = i;//將父親設置為自己。 for(int i = 0; i < n; i++) { scanf("%d %d %d %d", &data[i].id, &data[i].fid, &data[i].mid, &k); //直接讀進來，不使用中間變量。 //並沒有使用下標作為id啊。 visit[data[i].id] = true;//標記出現過了。 if(data[i].fid != -1) { visit[data[i].fid] = true; Union(data[i].fid, data[i].id);//將id作為並查集中的關鍵字合並。 } if(data[i].mid != -1) { visit[data[i].mid] = true; Union(data[i].mid, data[i].id); } for(int j = 0; j < k; j++) { scanf("%d", &data[i].cid[j]); visit[data[i].cid[j]] = true; Union(data[i].cid[j], data[i].id); } scanf("%d %d", &data[i].num, &data[i].area); } for(int i = 0; i < n; i++) { int id = find(data[i].id);//找到當前人的父親， ans[id].id = id;//現在這個ans中使用id作為下標索引了！ ans[id].num += data[i].num; ans[id].area += data[i].area; ans[id].flag = true; } for(int i = 0; i < 10000; i++) { if(visit[i])//如果它出現過。 i ans[find(i)].people++; if(ans[i].flag)//標記有幾簇人家。 cnt++; } for(int i = 0; i < 10000; i++) { if(ans[i].flag) { ans[i].num = (double)(ans[i].num * 1.0 / ans[i].people); ans[i].area = (double)(ans[i].area * 1.0 / ans[i].people); //並沒有多個num屬性，都是用一個存的，一開始就設為double。 //我居然還分開定義了。 } } sort(ans, ans + 10000, cmp1); printf("%d\n", cnt); for(int i = 0; i < cnt; i++) printf("%04d %d %.3f %.3f\n", ans[i].id, ans[i].people, ans[i].num, ans[i].area); return 0; }
1.將Father初始化為了自己，那麽find函數就簡單了，學習了
2.使用一個bool數組來標記出現過的點和沒出現過的點，就解決了一家只有一口的情況。
3.在求ans向量的時候，仍使用了find，其實不用map的，find找到父親都是可以用的。
4.因為data[i].id裏存的是已經出現過的id,對那些沒出現過的ans.flag肯定不會被標記為true的！
這是錯誤理解：ans裏存儲的所有的10000個人的情況，對於那些沒有出現過的id，ans[i].flag也是true，只不過它所有的數據都是0，在經過排序之後，再通過簇進行控制輸出，其他的不輸出。
總之，學習了。
PAT 1114 Family Property[並查集][難]