Greedy Gift Givers

This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

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 some or all of the other friends (although some might be cheap and give to no one). Likewise, each friend might or might not receive money from any or all of the other friends. Your goal is to deduce how much more money each person receives than they give.

The rules for gift-giving are potentially different than you might expect. Each person goes to the bank (or any other source of money) to get 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 7 among 2 friends would be 3 each for the friends with 1 left over – that 1 left over goes into the giver’s “account”. All the participants’ gift accounts start at 0 and are decreased by money given and increased by money received.

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 money each person ends up with.

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 characters, ‘\n’ and ‘\r’. Do not let your program get trapped by this!

PROGRAM NAME: gift1

INPUT FORMAT

Line #

Contents

A single integer, NP

2..NP+1

Line i+1 contains the name of group member i

NP+2..end

NP groups of lines organized like this:

The first line of each group tells the person's name who will be giving gifts.

The second line in the group contains two numbers:

The amount of money (in the range 0..2000) to be divided into gifts by the giver
NG_i (1 ≤ NG_i ≤ NP), the number of people to whom the giver will give gifts

If NG_i is nonzero, each of the next NG_i lines lists the name of a recipient of a gift; recipients are not repeated in a single giver's list.

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

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 starting 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 OUTPUT (file gift1.out)

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

OUTPUT EXPLANATION

dave	laura	owen	vick	amr
0	0	0	0	0

First, ‘dave’ splits 200 among ‘laura’, ‘owen’, and ‘vick’. That comes to 66 each, with 2 left over

-200+2	+66	+66	+66	0
-198	66	66	66	0

Second, ‘owen’ gives 500 to ‘dave’:

-198 +500	66	66 -500	66	0
302	66	-434	66	0

Third, ‘amr’ splits 150 between ‘vick’ and ‘owen’:

302	66	-434 +75	66 +75	-150
302	66	-359	141	-150

Fourth, ‘laura’ splits 0 between ‘amr’ and ‘vick’; no changes:

302

-359

141

-150

Finally, ‘vick’ gives 0 to no one:

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

CODE

Java

C++

Pascal

ANALYSIS

Russ Cox

The hardest part about this problem is dealing with the strings representing people’s names.

We keep an array of Person structures that contain their name and how much money they give/get.

The heart of the program is the lookup() function that, given a person’s name, returns their Person structure. We add new people with addperson().

Note that we assume names are reasonably short.

#include <stdio.h>
#include <string.h>
#include <assert.h>

#define MAXPEOPLE 10
#define NAMELEN	32

typedef struct Person Person;
struct Person {
    char name[NAMELEN];
    int total;
};

Person people[MAXPEOPLE];
int npeople;

void
addperson(char *name)
{
    assert(npeople < MAXPEOPLE);
	strcpy(people[npeople].name, name);
    npeople++;
}

Person*
lookup(char *name)
{
    int i;

    /* look for name in people table */
    for(i=0; i<npeople; i++)
	if(strcmp(name, people[i].name) == 0)
	    return &people[i];

    assert(0);	/* should have found name */
}

int
main(void)
{
    char name[NAMELEN];
    FILE *fin, *fout;
    int i, j, np, amt, ng;
    Person *giver, *receiver;

    fin = fopen("gift1.in", "r");
    fout = fopen("gift1.out", "w");

    fscanf(fin, "%d", &np);
    assert(np <= MAXPEOPLE);

    for(i=0; i<np; i++) {
	fscanf(fin, "%s", name);
	addperson(name);
    }

    /* process gift lines */
    for(i=0; i<np; i++) {
	fscanf(fin, "%s %d %d", name, &amt, &ng);
	giver = lookup(name);

	for(j=0; j<ng; j++) {
	    fscanf(fin, "%s", name);
	    receiver = lookup(name);
	    giver->total -= amt/ng;
	    receiver->total += amt/ng;
	}
    }

    /* print gift totals */
    for(i=0; i<np; i++)
	fprintf(fout, "%s %d\n", people[i].name, people[i].total);
    exit (0);
}