hdu1059(多重背包)

Dividing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19495    Accepted Submission(s): 5481

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.

 

Source

Mid-Central European Regional Contest 1999

 

Recommend

题目描述：求重量为1，2，3，4，5，6的石子，每个石子有a[i]个，能否将总重量分为两半。

明显的多重背包，重量和体积相等，背包容量为总重量的一半。

// 少打换行 ans/=2写早了 数组开小了 

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

#define maxn 60100

#define LL long long

using namespace std;

int a[10];

int d[maxn];

int ans;

void init()

{

  memset(d,0,sizeof(d));

}

void CompletePack(int volume,int weight)

{

    for(int i=volume;i<=ans;i++)

     d[i]=max(d[i],d[i-volume]+weight);

}

void ZeroPack(int volume,int weight)

{

    for(int i=ans;i>=volume;i--)

     d[i]=max(d[i],d[i-volume]+weight);

}

void MultiplePack(int volume,int num)

{



    if(volume*num>=ans)

      CompletePack(volume,volume);

    else

      {

        int k=1;

        while(k<=num)

        {

          ZeroPack(k*volume,k*volume);

          num-=k;

          k*=2;

        }

          ZeroPack(num*volume,num*volume);

      }

}

void solve()

{

    for(int i=1;i<=6;i++)

    {

          if(a[i]!=0)

          MultiplePack(i,a[i]);

    }



    //   for(int i=1;i<=ans;i++)

       // printf("%d %d\n",i,d[i]);



}

int main()

{

    int _case=0;

    while(~scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6]) )

    {

        init();

        if(a[1]==0 && a[2]==0 && a[3]==0 && a[4]==0 && a[5]==0 && a[6]==0)

        break;

          ans=(a[1]*1+a[2]*2+a[3]*3+a[4]*4+a[5]*5+a[6]*6);

        if((ans%2!=0))

            printf("Collection #%d:\nCan't be divided.\n\n",++_case);

        else

        {

            ans/=2;

             solve();

            if(d[ans]==ans)

            printf("Collection #%d:\nCan be divided.\n\n",++_case);

            else

            printf("Collection #%d:\nCan't be divided.\n\n",++_case);

        }

    }

  return 0;

}