【解題】Greedy Gift Givers

USACO Section 1.1 - Greedy Gift Givers


The Problem

A group of NP (2 ≤ NP ≤ 10) uniquely named friends has decided to exchange gifts of money. Each of these friends might or might not give some money to any or all of the other friends. Likewise, each friend might or might not receive money from any or all of the other friends. Your goal in this problem is to deduce how much more money each person gives than they receive.

The rules for gift-giving are potentially different than you might expect. Each person sets aside a certain amount of money to give and divides this money evenly among all those to whom he or she is giving a gift. No fractional money is available, so dividing 3 among 2 friends would be 1 each for the friends with 1 left over -- that 1 left over stays in the giver's "account".

In any group of friends, some people are more giving than others (or at least may have more acquaintances) and some people have more money than others.

Given a group of friends, no one of whom has a name longer than 14 characters, the money each person in the group spends on gifts, and a (sub)list of friends to whom each person gives gifts, determine how much more (or less) each person in the group gives than they receive.


Important Note

The grader machine is a Linux machine that uses standard Unix conventions: end of line is a single character often known as '\n'. This differs from Windows, which ends lines with two charcters, '\n' and '\r'. Do not let your program get trapped by this!


Program Name: gift1


Input Format

Line 1: The single integer, NP

Lines 2..NP+1: Each line contains the name of a group member

Lines NP+2..end: NP groups of lines organized like this:
The first line in the group tells the person's name who will be giving gifts.
The second line in the group contains two numbers: The initial amount of money (in the range 0..2000) to be divided up into gifts by the giver and then the number of people to whom the giver will give gifts, NG_i (0 ≤ NG_i ≤ NP-1).
If NG_i is nonzero, each of the next NG_i lines lists the the name of a recipient of a gift.


Output Format

The output is NP lines, each with the name of a person followed by a single blank followed by the net gain or loss (final_money_value - initial_money_value) for that person. The names should be printed in the same order they appear on line 2 of the input. All gifts are integers. Each person gives the same integer amount of money to each friend to whom any money is given, and gives as much as possible that meets this constraint. Any money not given is kept by the giver.


Sample Input (file gift1.in)

5
dave
laura
owen
vick
amr
dave
200 3
laura
owen
vick
owen
500 1
dave
amr
150 2
vick
owen
laura
0 2
amr
vick
vick
0 0


Sample Output (file gift1.out)

dave 302
laura 66
owen -359
vick 141
amr -150


解題思考

　　首先，我宣告了一個包含了「姓名」(name)與「收支金額」(money)的結構體。在讀入所有人的名字，存入結構體中的 name 成員之後，同時將其 money 成員歸零(在起初，並未有任何收入或支出)。

　　接著讀入所有的「送禮紀錄」之後，則將送禮者的 money 成員減去送禮花費的金額。但是要注意，由於送禮的金額是「盡可能多給」，且能夠「平分給所有人」的金額。所以實際花費的金額，應該是每個人的「持有金額」減去「此金額除以收禮者人數的餘數」。

　　然後將每個收禮者的 money 成員加上收到的禮物金額，也就是「送禮金額」除以「收禮人數」。

　　讀完所有送禮紀錄後，直接將每個人的 name 與 money 成員成對輸出，就是題目要求的結果了。


參考解答(C++)

#include <fstream>
#include <cstring>

using namespace std;

struct person
{
    char name[13];
    int money;
};

int main(void)
{
    ifstream in("gift1.in", ios::in);
    ofstream out("gift1.out", ios::out);

    int NP;
    in >> NP;

    person *people = new person[NP];
    for (int i = 0; i < NP; i++)
    {
        in >> people[i].name;

        people[i].money = 0;
    }

    for (int i = 0; i < NP; i++)
    {
        char name[13];
        int money, number;
        in >> name >> money >> number;

        // 忽略不送任何禮物的情況
        if (money && number)
        {
            for (int j = 0; j < NP; j++)
            {
                if (!strcmp(people[j].name, name))
                {
                    // 將自身金額扣除支出金額
                    people[j].money -= money - money % number;
                    break;
                }
            }
        }

        for (int j = 0; j < number; j++)
        {
            char recipient[13];
            in >> recipient;

            // 將禮物收受者金額加上禮物金額
            for (int k = 0; k < NP; k++)
            {
                if (!strcmp(people[k].name, recipient))
                {
                    people[k].money += money / number;
                    break;
                }
            }
        }
    }

    in.close();

    for (int i = 0; i < NP; i++)
    {
        out << people[i].name << " " << people[i].money << endl;
    }

    out.close();

    return 0;
}