HDU Dividing

Dividing

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 113 Accepted Submission(s): 50

Problem Description

Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.

 

Input

Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0''. The maximum total number of marbles will be 20000.

The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.

 

Output

For each colletcion, output ``Collection #k:'', where k is the number of the test case, and then either ``Can be divided.'' or ``Can't be divided.''.

Output a blank line after each test case.

 

Sample Input

1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0

 

Sample Output

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

分析：

 这是一个典型的多重背包问题，其算法伪代码为：

procedure MultiplePack(cost,weight,amount)

    if cost*amount>=V

        CompletePack(cost,weight)

        return

    integer k=1

    while k<amount

        ZeroOnePack(k*cost,k*weight)

        amount=amount-k

        k=k*2

    ZeroOnePack(amount*cost,amount*weight)

虑二进制的思想，我们考虑把第i种物品换成若干件物品，使得原问题中第i种物品可取的每种策略——取0..n[i]件——均能等价于取若干件代换以后的物品。另外，取超过n[i]件的策略必不能出现。

方法是：将第i种物品分成若干件物品，其中每件物品有一个系数，这件物品的费用和价值均是原来的费用和价值乘以这个系数。使这些系数分别为1,2,4,...,2^(k-1),n[i]-2^k+1，且k是满足n[i]-2^k+1>0的最大整数。例如，如果n[i]为13，就将这种物品分成系数分别为1,2,4,6的四件物品。

 

 

 

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int mar[7],f[120005],v,k;
//01背包
void ZeroOnePack(int va,int c)//va为价值，c为花费
{
    for(int i = v ; i >= va; i--)
    if(f[i-va]+c>f[i])
    f[i] = f[i-va]+c;
}
//完全背包
void complatePack(int va,int c)//va为价值，c为花费
{
    for(int i = va; i <= v; i++)
    if(f[i-va]+c>f[i])
    f[i] = f[i-va]+c;
}
//多重背包
void mutilPack(int va,int c,int n)//va为价值，c为花费，n为数量
{
    if(n*va>=v)
    complatePack(va,c);
    else
    {
       int count = 1;
        while(count<n)
        {
            ZeroOnePack(count*va,count*c);
            n -=count;
            count =2*count;//关键：这里把物品用2进制分类
        }
        ZeroOnePack(n*va,n*c);
    }
}
int main()
{
    int total,num=1;
    while(1)
    {
        total = 0;
        for(int i = 1; i<7;i++)
        {
            scanf("%d",&mar[i]);
            total += i*mar[i];
        }
        if(total==0)break;
        if(total%2)
        {
            printf("Collection #%d:\nCan't be divided.\n\n",num);
            num++;
            continue;
        }
        v = total/2;
        memset(f,0,sizeof(f));
        for(int i=1;i<7;i++)
        mutilPack(i,i,mar[i]);
        if(f[v]==v) printf("Collection #%d:\nCan be divided.\n\n",num);
        else printf("Collection #%d:\nCan't be divided.\n\n",num);
        num++;
    }
    return 0;
}