• pat甲级1114


    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 N lines follow, each gives the infomation of a person who owns estate in the format:

    ID Father Mother 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 (0k5) 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

    注意ID没有说不等于0.因此判断时为大于等于0,而不是大于。否则测试点2、4会错误。

    我用的深度优先搜索求联通分量来完成的这道题,另外用并查集好像也比较方便。

    由于ID不连续,所以求连通分量之前先将所有ID放到vector里面,而且用邻接表存储图更节省空间和时间。

     1 #include <iostream>
     2 #include <vector>
     3 #include <algorithm>
     4 using namespace std;
     5 
     6 bool marked[10000];
     7 vector<vector<int>> G(10000);
     8 int estate[10000], area[10000];
     9 
    10 struct family
    11 {
    12     int id = 10000, size = 0;
    13     double estate = 0.0, area = 0.0;
    14 };
    15 
    16 void connect(int id)
    17 {
    18     int t;
    19     cin >> t;
    20     if (t >= 0)
    21     {
    22         G[id].push_back(t);
    23         G[t].push_back(id);
    24     }
    25 }
    26 
    27 void dfs(int s, family& f)
    28 {
    29     marked[s] = true;
    30     f.size++;
    31     if (s < f.id) f.id = s;
    32     f.area += area[s];
    33     f.estate += estate[s];
    34     for (int w : G[s])
    35         if (!marked[w]) dfs(w, f);
    36 }
    37 
    38 bool cmp(family a, family b)
    39 {
    40     if (a.area != b.area) return a.area > b.area;
    41     else return a.id < b.id;
    42 }
    43 
    44 int main()
    45 {
    46     int N;
    47     cin >> N;
    48     int i, j, id, k;
    49     vector<int> allId;
    50     for (i = 0; i < N; i++)
    51     {
    52         cin >> id;
    53         allId.push_back(id);
    54         for (j = 0; j < 2; j++)
    55             connect(id);
    56         cin >> k;
    57         for (j = 0; j < k; j++)
    58             connect(id);
    59         cin >> estate[id] >> area[id];
    60     }
    61     vector<family> v;
    62     for (int id : allId)
    63     {
    64         if (!marked[id])
    65         {
    66             family f;
    67             dfs(id, f);
    68             f.area /= f.size;
    69             f.estate /= f.size;
    70             v.push_back(f);
    71         }
    72     }
    73     sort(v.begin(), v.end(), cmp);
    74     cout << v.size() << endl;
    75     for (i = 0; i < v.size(); i++)
    76         printf("%04d %d %.3f %.3f
    ", v[i].id, v[i].size, v[i].estate, v[i].area);
    77     return 0;
    78 }
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  • 原文地址:https://www.cnblogs.com/lxc1910/p/9525103.html
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