usaco Greedy Gift Givers 报告

题意：对于一群（NP个）要互送礼物的朋友，GY要确定每个人送出的钱比收到的多多少。 在这一个问题中，每个人都准备了一些钱来送礼物，而这些钱将会被平均分给那些将收到他的礼物的人。 然而，在任何一群朋友中，有些人将送出较多的礼物(可能是因为有较多的朋友)，有些人有准备了较多的钱。 给出一群朋友，没有人的名字会长于 14 字符，给出每个人将花在送礼上的钱，和将收到他的礼物的人的列表， 请确定每个人收到的比送出的钱多的数目。

题解：就是每个人的钱相互分，算一下

代码：

/*
ID:   lishicao
PROG: gift1
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
using namespace std ;

ifstream fin  ( "gift1.in"  ) ;
ofstream fout ( "gift1.out" ) ;

int  main()
{
    string  name[15] ;
    string  temp ;
    int  N , i , j , k , cnt , Gift , num ;
    int  money[15] ;

    memset( money , 0 , sizeof( money ) ) ;

    fin >> N ;
    for( i = 0 ; i < N ; i ++ )
        fin >> name[i] ;
    for( i = 0 ; i < N ; i ++ )
    {
        fin >> temp ;
        for( j = 0 ; j < N ; j ++ )
            if( temp == name[j] ) break ;
        fin >> Gift >> num ;
        if( num > 0 ) {
            Gift = ( Gift / num ) * num ;
            money[j] -= Gift ;
        }
        for( k = 0 ; k < num ; k ++ )
        {
            fin >> temp ;
            for( cnt = 0 ; cnt < N ; cnt ++ )
                if( temp == name[cnt] ) break ;
            money[cnt] += Gift / num ;
        }
    }
    for( i = 0 ; i < N ; i ++ )
        fout << name[i] << " " << money[i] << endl ;
    return 0 ;
}

官方题解：

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);
}