Dividing(多重背包、单调队列优化dp)

Dividing

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 0   Accepted Submission(s) : 0

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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.

题意：一共有6种物品，价值分别是1,2,3,4,5,6,给出每种物品的数目，判断师傅能够平分

思路：典型的多重背包，防止超时要进行一些必要的剪枝和二进制优化

多重背包二进制拆分实现:

跟完全背包一样的道理，利用二进制的思想将n[i]件物品i拆分成若干件物品，目的是在0-n[i]中的任何数字都能用这若干件物品代换，另外，超过n[i]件的策略是不允许的。

方法是将物品i分成若干件，其中每一件物品都有一个系数，这件物品的费用和价值都是原来的费用和价值乘以这个系数，使得这些系数分别为1,2,4,…,2^(k-1),n[i]-2^k+1，且k满足n[i]-2^k+1>0的最大整数。例如，n[i]=13，就将该物品拆成系数为1、2、4、6的四件物品。分成的这几件物品的系数和为n[i]，表明不可能取多于n[i]件的第i种物品。另外这种方法也能保证对于0..n[i]间的每一个整数，均可以用若干个系数的和表示。

 

代码如下：

#include 
#include 
#include 
using namespace std;

int dp[100000];

int main()
{
    int a[11],sum,v,i,j,k,cnt,cas = 1;
    while(~scanf("%d",&a[1]))
    {
        sum = a[1];
        for(i = 2;i<=6;i++)
        {
            scanf("%d",&a[i]);
            sum+=i*a[i];
        }
        if(!sum)
        break;
        printf("Collection #%d:\n",cas++);
        if(sum%2)//总和为奇数，必定不能平分
        {
            printf("Can't be divided.\n\n");
            continue;
        }
        v = sum/2;
        memset(dp,0,sizeof(dp));
        dp[0] = 1;
        for(i = 1;i<=6;i++)
        {
            if(!a[i])
            continue;
            for(j = 1;j<=a[i];j*=2)//二进制优化
            {
                cnt = j*i;
                for(k = v;k>=cnt;k--)
                {
                    if(dp[k-cnt])//必须前面的能够放入背包，现在的才能放入背包
                    dp[k] = 1;
                }
                a[i]-=j;
            }
            cnt = a[i]*i;//剩下的
            if(cnt)
            {
                for(k = v;k>=cnt;k--)
                {
                    if(dp[k-cnt])
                    dp[k] = 1;
                }
            }
        }
        if(dp[v])
        printf("Can be divided.\n\n");
        else
        printf("Can't be divided.\n\n");
    }

    return 0;
}