USACO打怪升级 (六): Greedy Gift Givers

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.

题目大意： 
   一群名字唯一的人，决定交换钱作为礼物，每个人会把一些钱给其他任何人，同样，每个人也可能受到来自其他任何人的钱。我们的问题就是计算每个人给出去的钱比他收到的钱多多少？
    发钱的规则是每个人拿出一定数量的钱，来平均分给他想送的人。分出来的钱不会有分数，例如，把3元钱分2两个人，每个人都是一元，还剩余一元就留给自己。
    给定一群人，同时列出每个人将要送的钱的总额，和他要送给哪些人，计算最后每个人发出去的钱比收到的钱多多少？

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, NGi (0 ≤ NGi ≤ NP-1).
If NGi is nonzero, each of the next NGi 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.

分析：
  这种问题，只要小心读题目，很快就能想到办法。我们只要以每个人为单位进行计算，不需要什么特定办法。

下面是我自己编写的代码：

/*
 ID: fuchenc1
 PROG: gift1
 LANG: C++ 
*/

#include <iostream>
#include <map>
#include <fstream>
#include <string>
using namespace std;

int Solution() 
{
    ifstream fin("gift1.in");
    ofstream fout("gift1.out");
    const int NP_MAX=10+2;
    int NP;
	fin>>NP;
	map<string,int> mapPersonGifts;
	string arrayRecord[NP_MAX]; 
	for(int i=0;i<NP;i++)
	{
		string strPersonName;
		fin>>strPersonName;
		arrayRecord[i]=strPersonName;
		mapPersonGifts[strPersonName]=0;
	}

	for(int i=0;i<NP;i++)
	{
	   int nMoney,nPersonCnt;
	   string strGiverName;
       fin>>strGiverName>>nMoney>>nPersonCnt;
       int aveMoney=0;
       if(nPersonCnt)
       {
  		  aveMoney=nMoney/nPersonCnt;
       }
       for(int j=0;j<nPersonCnt;j++)
	   {
	     string strReciverName;
	     fin>>strReciverName;
	     mapPersonGifts[strReciverName]+=aveMoney;	
	   }
	   mapPersonGifts[strGiverName]+=(-aveMoney*nPersonCnt);
	}
	map<string ,int>::const_iterator map_it=mapPersonGifts.begin();
	for(int i=0;i<NP;i++)
	{
		fout<<arrayRecord[i]<<" "<<mapPersonGifts[arrayRecord[i]]<<endl; 
	}
	return 0;
}
int main()
{
    return Solution();
    
} 

  因为采用了map容器，所以处理起来比较方便，但是需要注意的是题目要求按照名字输入顺序输出每个人的礼物情况，而map容器对元素的排序是根据元素的key类型进行排序，所以如果直接遍历map容器进行输出，就会是按照名字的升序输出，所以答案会出错，所以我另外用了一个数组记录下输入名字的顺序。

下面是通过测试后，网站给出的参考答案：

#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);
}

标准答案也非常容易看懂，但是如果直接采用map容器，感觉处理起来更方便。