USACO . Greedy Gift Givers

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.

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, 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.

 

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 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

 

 

 

 

    简单水题，大意是有 NP 个朋友交换礼物（钱），输入第一行为 NP，接着 NP 行是朋友们的名字，再然后就是 NP 组数据，每组第一行是分发礼物的人的名字，第二行是他的初始资金和要分给 NG 个人，接着 NG 行是要分给的人名。

    注意题目说了钱要分的平均（evenly）且不能有小数（no fractional），因此我的做法是，在读入 NP 个名字时金钱初始化为 0，读入 NP 组数据时先减去他的初始金额，再加回（初始金额 % NG），这样分给其他人的礼物金额就是（初始金额 / NG）。

    难点主要在于字符串的处理，我偷懒用了 map，为了效率 key 类型为 char* 而不是 string，这样就需要自己重载操作符。跟直接使用 struct 比较，也没什么优势，偷懒的代价。

#include <iostream>
#include <fstream>
#include <cstring>
#include <map>
using namespace std;
#define Native 0
#if Native
    #define fin cin
    #define fout cout
#else
    ifstream fin("gift1.in");
    ofstream fout("gift1.out");
#endif
struct ptrCmp{
    bool operator()(const char*s1,const char*s2)const{
        return strcmp(s1,s2)<0;
    }
};
map<char*,int,ptrCmp> mp;
map<char*,int,ptrCmp>::iterator it;
int main(){
    int NP,NG,init,gift;
    char tname[20],fri[10][20];

    fin>>NP;
    for(int i=0;i<NP;i++){
        fin>>fri[i];
        mp[fri[i]]=0;
    }
    for(int i=0;i<NP;i++){
        fin>>tname>>init>>NG;
        mp[tname]-=init;
        if(!NG) continue;
        //题目数据良心，只有 0 0，如果出 500 0 之类的我这里就挂了
        //上面两行代码应该交换下，这样应该就没问题了
        mp[tname]+=init%NG;
        gift=init/NG;
        while(NG--){
            fin>>tname;
            mp[tname]+=gift;
        }
    }
    for(int i=0;i<NP;i++)
        fout<<fri[i]
            <<' '
            <<mp[fri[i]]
            <<endl;
    return 0;
}

    
好吧，官方代码再次把我吊打了，简明易懂很优美有木有！来欣赏下吧~

#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);
}

 

By Black Storm（使用为知笔记）