PKU-1014 Dividing (多重背包)

最新推荐文章于 2019-01-08 01:51:44 发布

gk1988

最新推荐文章于 2019-01-08 01:51:44 发布

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本文探讨如何使用多重背包问题解决公平分配不同价值物品的挑战。通过输入描述的物品数量和价值，输出是否可以公平分配给两个接收者，使得每个接收者的总价值相等。该文提供了一个算法解决方案，并通过实例演示了实现过程。

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原题链接 http://poj.org/problem?id=1014

Dividing

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 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.

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.

Source Code

/*
此题可以用多重背包问题方法求解，其中包的容量为总量的一半，
各个物品的cost为i, weight为1, 且该物品有count[i]个，要求完全装满
*/  

#include <iostream>
using namespace std;

#define INF 1000000

//V指背包的最大容量
int result[60004],V;

//01背包
void ZeroOnePack(int cost, int weight)
{
	for(int i = V; i >= cost; i--)
		if(result[i] < result[i-cost]+weight)
			result[i] = result[i-cost]+weight;
}

//完全背包
void CompletePack(int cost, int weight)
{
	for(int i = cost; i <= V; i++)
		if(result[i] < result[i-cost]+weight)
			result[i] = result[i-cost]+weight;
}

//多重背包
void MultiplePack(int cost, int weight, int count)
{
	if(cost * count >= V)
		CompletePack(cost,weight);
	else
	{
		//采进二进制思想压缩，转化为01背包(13=1+2+4+6,即1,2,4，……，count-2^k+1)
		int k = 1;
		while(k < count)
		{
			ZeroOnePack(k*cost, k*weight);
			count -= k;
			k*=2;
		}
		ZeroOnePack(count*cost, count*weight);
	}
	
}


int main()
{
	int count[10];
	int i, collection;
	collection = 1;
	while(scanf("%d %d %d %d %d %d",&count[1], &count[2], 
		&count[3], &count[4], &count[5], &count[6])!=EOF)
	{
		if (count[1] == 0 && count[2] == 0 && count[3] == 0 &&
			count[4] == 0 && count[5] == 0 && count[6] == 0){
			break;
		}		
		
		printf("Collection #%d:\n", collection ++);

		//要求完全放满，初始化result[0]为0，其它全为负无穷
		for (i = 0; i < 60004; i++) {
			result[i] = -1 * INF;
		}
		result[0] = 0;
		V = 0;

		for (i = 1; i <= 6; i++) {
			V += (i * count[i]);
		}

		//总量是奇数不可分
		if (V % 2 == 1) {
			printf("Can't be divided.\n\n");
			continue;
		}

		V = V / 2;

		for(i = 1; i <= 6 ; i++)
			MultiplePack(i, 1, count[i]);
			
		if (result[V] < 0) {
			printf("Can't be divided.\n\n");
		} else {
			printf("Can be divided.\n\n");
		}	
	}
	
}