POJ_1014 Dividing

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最新推荐文章于 2022-04-02 21:33:34 发布

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本文链接：https://blog.youkuaiyun.com/Galaxy_Engineer/article/details/119577528

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马莎和比尔有一堆大理石要平分。他们希望将收藏品分成两半，使得两人获得相等价值的大理石。但由于一些大理石的价值不同，他们决定为每个大理石分配一个1到6之间的数值。然后他们需要找到一种方式，使得每人都能获得相同总价值的大理石。题目要求编写程序检查是否存在公平的分割方法。输入包含多个测试用例，每个用例表示一堆大理石，给出每种价值大理石的数量。输出应表明每个测试用例是否可以公平分割。

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Dividing

链接

POJ_1014 Dividing

Describe

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 file describes one collection of marbles to be divided. The lines contain six non-negative integers $n_1 , . . . , n_6$ , 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 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.

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.

解析

本题可以通过二进制优化将多重背包转化为01背包。

若某一多重背包的备选物品的可选数量为 $n_i$ ，价值为 $v_i$ ，重量为 $w_i$ ，我们可以将 $n_i$ 分解成 $1,2,4,8,...,2^k$ 与 $n_i-2^k$ ，把这些数分别作为 $v_i$ 和 $w_i$ 的系数，重新定义为01背包的备选物品。

代码

#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int a[11],k,tot,v,f[1000001],test;
bool flag;
void pack(int wei,int val)
{
	for(int j=v;j>=wei;j--)
	{
		f[j]=max(f[j],f[j-wei]+val);
		if(f[j]==v)
		{
			flag=true;
			break;
		}
	}
}
int main()
{
	while(1)
	{
		test++;
		tot=0;
		for(int i=1;i<=6;i++)
		{
			scanf("%d",&a[i]);
			tot+=a[i]*i;
		}
		if(!tot)
			break;
		if(tot&1)
		{
			printf("Collection #%d:\nCan't be divided.\n\n",test);
			continue;
		}
		v=tot/2;
		flag=false;
		memset(f,-1,sizeof(f));
		f[0]=0;
		for(int i=1;i<=6;i++)
		{
			int k=1,num=a[i];
			while(num-k>0)
			{
				pack(k*i,k*i);
				num=num-k;
				k=k<<1;
				if(flag)
					break;
			}
			pack(num*i,num*i);
			if(flag)
				break;
		}
		if(flag)
			printf("Collection #%d:\nCan be divided.\n\n",test);
		else
			printf("Collection #%d:\nCan't be divided.\n\n",test);
	}
	return 0;
}