PTA甲级 1114 Family Property (25 point(s))

强烈推荐,刷PTA的朋友都认识一下柳神–PTA解法大佬

本文由参考于柳神博客写成

柳神的CSDN博客,这个可以搜索文章

柳神的个人博客,这个没有广告,但是不能搜索

还有就是非常非常有用的算法笔记全名是

算法笔记  上级训练实战指南       //这本都是PTA的题解
算法笔记

PS 今天也要加油鸭

题目原文

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* *C**h**i**l**d*1⋯*C**h**i**l**d**k* *M**e**s**t**a**t**e* `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**i’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

生词如下:

PS:这题就属于我看不懂的题目了.就都不懂.

肯定是和家庭收入有关.还有房子啥的.

肯定要用到树.这题算的东西不多.需要用到的主要也是树的遍历.

题目大意:

就是给你这样N行数据

格式为

你的ID 你爸的ID 你妈的ID K个孩子(K可以为0) 你的房产的数量房产一共有多少面积

然后让你算出一共有几个家庭.

每个家庭一共有多少人.每个人平均能分到多少房子平均能分到多少面积.

不会写

柳神思路:

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～～

Copy柳神代码

#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;elsereturn 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;
}