poj 1014(多重背包讲解）

问题描述

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.

输入

Each line in the input file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , 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.

输出

For each collection, 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.

样例输入

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

样例输出

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

首先是普通dp的形式，只需判断最后dp[sum/2]==sum/2?即可

代码如下：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int dp[200002];
int num[7];
int sum;int cas;

bool init()
{


    sum=0;
    bool flag=false;
    for(int i=1;i<=6;i++)
    {
         scanf("%d",&num[i]);
         if(num[i]!=0)flag=true;
         sum+=i*num[i];
    }
    return flag;

}

int main()
{
    while(init())
    {
        printf("Collection #%d:\n", ++cas);
        if(sum%2==1)
        {
            puts("Can't be divided.\n");
            continue;
        }
        for(int i=1;i<=6;i++)
           {
               if(num[i]*i>=sum/2)
               {
                   for(int j=i;j<=sum/2;j++)
                    dp[j]=max(dp[j-i]+i,dp[j]);
               }
               else
               {
                   int k=1;int M=num[i];
                   //利用位二进制原理
                   while(k<M)
                   {
                       for(int j=sum/2;j>=k*i;j--)
                        dp[j]=max(dp[j],dp[j-k*i]+k*i);

                       M-=k;k=k*2;
                   }
                   for(int j=sum/2;j>=i*M;j--)
                    dp[j]=max(dp[j],dp[j-M*i]+M*i);
               }
           }
        if(dp[sum/2]!=sum/2)
        {

            puts("Can't be divided.\n");
        
            continue;
        }
        else
        {

            puts("Can be divided.\n");

            continue;
        }

    }
}

还有一种就是O(VN)的方法，因为是背包可行性问题，所以只需要记录可不可以就行了。

代码如下：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int dp[200002],use[200002];//dp[i]表示是否能沾满i个单位空间，use[i]表示第i种用了多少次.
int num[7];
int sum;int cas;

bool init()
{
    sum=0;
    bool flag=false;
    for(int i=1;i<=6;i++)
    {
         scanf("%d",&num[i]);
         if(num[i]!=0)flag=true;
         sum+=i*num[i];
    }
    return flag;

}

int main()
{
    while(init())
    {
        memset(dp,0,sizeof(dp));
        printf("Collection #%d:\n", ++cas);
        if(sum%2==1)
        {
            puts("Can't be divided.\n");
            continue;
        }
        dp[0]=1;
        for(int i=1;i<=6;i++)
           {
               memset(use,0,sizeof(use));
               for(int j=i;j<=sum/2;j++)
               {
                   if(!dp[j]&&dp[j-i]&&use[j-i]<num[i])
                   {//这里解释一下，如果dp[j]已经不为0了，那么就没必要再算dp[j]了，因为已经能填满j个空间了
                   // dp[j-i]是看j-i有没有被填满过，如果连j-i都没有被填满过，那么当前是肯定 填不了j的。
                   //use[j-i]<num[i]判断是不是已经把东西用光了。
                      dp[j]=1;
                       use[j]=use[j-i]+1;
                   }
               }
           }
        if(!dp[sum/2])
        {

            puts("Can't be divided.\n");

            continue;
        }
        else
        {

            puts("Can be divided.\n");

            continue;
        }

    }
}