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 … Childk 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; Childi’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
题目大意:给定每个人的家庭成员和其自己名下的房产,请你统计出每个家庭的人口数、人均房产面积及房产套数。首先在第一行输出家庭个数(所有有亲属关系的人都属于同一个家庭)。随后按下列格式输出每个家庭的信息:家庭成员的最小编号 家庭人口数 人均房产套数 人均房产面积。其中人均值要求保留小数点后3位。家庭信息首先按人均面积降序输出,若有并列,则按成员编号的升序输出。
分析:用并查集。分别用两个结构体数组,一个data用来接收数据,接收的时候顺便实现了并查集的操作union,另一个数组ans用来输出最后的答案,因为要计算家庭人数,所以用visit标记所有出现过的结点,对于每个结点的父结点,people++统计人数。标记flag == true,计算true的个数cnt就可以知道一共有多少个家庭。排序后输出前cnt个就是所求答案~~
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#include <cstdio> #include <algorithm> using namespace std; struct DATA { int id, fid, mid, num, area; int cid[10]; }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]) 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; for(int i = 0; i < n; i++) { scanf("%d %d %d %d", &data[i].id, &data[i].fid, &data[i].mid, &k); visit[data[i].id] = true; if(data[i].fid != -1) { visit[data[i].fid] = true; Union(data[i].fid, data[i].id); } 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); ans[id].id = id; 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]) ans[find(i)].people++; if(ans[i].flag) 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); } } 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; } |
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