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 Nlines 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≤k≤5) 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题目大意：找出一个家庭中人均有几套房产，和人均面积；输入中为 当前人的ID 父亲ID 母亲ID 当前人拥有几套 当前人拥有总面积。如果父母去世了，则用-1表示。输出要求按平均房产套数递减，如果相同，那么按ID递增排列。
//确实是使用并查集。
//遇到问题：遇到像8888这种单户的，该怎么处理，我写的里边似乎处理不了。
#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;//这里出现了问题啊，得判断是否等于，等于的话，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; //先将当前的人和父亲母亲合并 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]++; //怎么记录每个簇里有谁呢？ vt[peo[i].father].push_back(i); } } //最终还得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; }
//写成了这样，最终决定放弃！
代码转自：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个孩子。 }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])//这样真的好简便，我为啥写那么复杂呢。 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;//将父亲设置为自己。 for(int i = 0; i < n; i++) { scanf("%d %d %d %d", &data[i].id, &data[i].fid, &data[i].mid, &k); //直接读进来，不使用中间变量。 //并没有使用下标作为id啊。 visit[data[i].id] = true;//标记出现过了。 if(data[i].fid != -1) { visit[data[i].fid] = true; Union(data[i].fid, data[i].id);//将id作为并查集中的关键字合并。 } 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);//找到当前人的父亲， ans[id].id = id;//现在这个ans中使用id作为下标索引了！ 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])//如果它出现过。 i ans[find(i)].people++; if(ans[i].flag)//标记有几簇人家。 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); //并没有多个num属性，都是用一个存的，一开始就设为double。 //我居然还分开定义了。 } } 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.将Father初始化为了自己，那么find函数就简单了，学习了
2.使用一个bool数组来标记出现过的点和没出现过的点，就解决了一家只有一口的情况。
3.在求ans向量的时候，仍使用了find，其实不用map的，find找到父亲都是可以用的。
4.因为data[i].id里存的是已经出现过的id,对那些没出现过的ans.flag肯定不会被标记为true的！
这是错误理解：ans里存储的所有的10000个人的情况，对于那些没有出现过的id，ans[i].flag也是true，只不过它所有的数据都是0，在经过排序之后，再通过簇进行控制输出，其他的不输出。
总之，学习了。