单词积累
decimal 小数的，十进制的
descending 下降的
ascending 上升的，增长的
the total number of sets of the real estate under his/her name 其名下房产的总套数
题目
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 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; child'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
结尾无空行
Sample Output:
3
8888 1 1.000 1000.000
0001 15 0.600 100.000
5551 4 0.750 100.000
结尾无空行
思路
很典型的并查集题目，就是元素过多，处理起来较麻烦。比起在union过程中处理元素值，更简单的做法是合并完全部遍历计算。
此外还涉及到了排序，格式化输出等知识点。
最坑的点是默认id编号是从1开始的，居然导致了三个样例出错，检查了半天，牢记不要想当然，题目不指名，就是从0开始编。
代码
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10000;
int father[maxn];
int numbers[maxn];
int visit[maxn];
double sets[maxn];
double areas[maxn];
struct fam{
int id;
int number;
double sets;
double areas;
}Family[maxn];
bool cmp(fam a, fam b) {
if (a.areas == b.areas) return a.id < b.id;
else return a.areas > b.areas;
}
void inital() {
for (int i = 0; i < maxn; i++) {
father[i] = i;
numbers[i] = 1;
visit[i] = 0;
sets[i] = 0;
areas[i] = 0;
}
return;
}
int findfather(int x) {
while (x != father[x]) {
x = father[x];
}
return x;
}
void Union(int a, int b) {
visit[a] = 1;
visit[b] = 1;
int x = findfather(a);
int y = findfather(b);
if (x == y) return ;
else if (x < y) {
father[y] = x;
} else if (x > y) {
father[x] = y;
}
return ;
}
int main() {
inital();
int N;
cin>>N;
for (int i = 0; i < N; i++) {
int id, idf, idm, idnc;
double estate, area;
cin>>id>>idf>>idm>>idnc;
if (idf != -1) Union(id, idf);
if (idm != -1) Union(id, idm);
int idc[6];
for (int j = 0; j < idnc; j++) {
cin>>idc[j];
Union(id, idc[j]);
}
cin>>estate>>area;
visit[id] = 1;
sets[id] += estate;
areas[id] += area;
}
for (int i = 1; i < maxn; i++) {
if(sets[i] != 0 || areas[i] != 0) {
int fa = findfather(i);
if (i != fa) {
sets[fa] += sets[i];
areas[fa] += areas[i];
}
}
}
int clus = 0;
for (int i = 0; i < maxn; i++) {
if(i != father[i]) {
int fa = findfather(i);
numbers[fa]++;
}
}
for (int i = 0; i < maxn; i++) {
if(i == father[i] && visit[i] != 0) {
Family[clus].id = i;
Family[clus].number = numbers[i];
Family[clus].sets = (double)(sets[i] * 1.0)/numbers[i];
Family[clus].areas= (double)(areas[i] * 1.0)/numbers[i];
clus++;
}
}
sort(Family, Family+clus, cmp);
cout<<clus<<endl;
for (int i = 0; i < clus; i++) {
printf("%04d %d %.3f %.3f\n", Family[i].id, Family[i].number, Family[i].sets, Family[i].areas);
}
}
