hdu-1059-多重背包-Dividing

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 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 01 0 0 0 1 10 0 0 0 0 0

Sample Output

Collection #1:Can't be divided.
Collection #2:Can be divided.

题意:有价值分别为1,2,3,4,5,6的弹珠,弹珠的数量分别为输入数据的n1,n2,n3,n4,n5,n6,问能否将其平均分成价值相等的两份

直接用多重背包写超时了,

f[i][v]=max{f[i-1][v],f[i-1][v-k*c[i]]+k*w[i]|0<=k<=n[i]}
复杂度是O(V*Σn[i])。
参考了别人的代码.
将其转化为0-1背包及完全背包问题

转化为01背包问题:把第i种物品换成n[i]件01背包中的物品，则得到了物品数为Σn[i]的01背包问题，直接求解，复杂度仍然是O(V*Σn[i])。

但是我们期望将它转化为01背包问题之后能够像完全背包一样降低复杂度。仍然考虑二进制的思想，我们考虑把第i种物品换成若干件物品，使得原问题中第i种物品可取的每种策略——取0..n[i]件——均能等价于取若干件代换以后的物品。另外，取超过n[i]件的策略必不能出现。

方法是：将第i种物品分成若干件物品，其中每件物品有一个系数，这件物品的费用和价值均是原来的费用和价值乘以这个系数。使这些系数分别为1,2,4,...,2^(k-1),n[i]-2^k+1，且k是满足n[i]-2^k+1>0的最大整数。例如，如果n[i]为13，就将这种物品分成系数分别为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，这个证明可以分0..2^k-1和2^k..n[i]两段来分别讨论得出，

这样就将第i种物品分成了O(log n[i])种物品，将原问题转化为了复杂度为O(V*Σlog n[i])的01背包问题，是很大的改进。

下面给出O(log amount)时间处理一件多重背包中物品的过程，其中amount表示物品的数量：

procedure MultiplePack(cost,weight,amount)

if cost*amount>=V

{

CompletePack(cost,weight)

Return

}

integer k=1

while k<amount

{

ZeroOnePack(k*cost,k*weight)

amount=amount-k

k=k*2

}

ZeroOnePack(amount*cost,amount*weight)

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int dp[200000];
int sum;
void cpl(int c,int w)//完全背包
{
int i;
for(i=c;i<=sum/2;i++)
{
if(dp[i]<dp[i-c]+w)
dp[i]=dp[i-c]+w;
}
}
void one(int c,int w)//0-1背包
{
int i;
for(i=sum/2;i>=c;i--)
{
if(dp[i]<dp[i-c]+w)
dp[i]=dp[i-c]+w;
}
}
void muty(int c,int w,int amount)//多重背包
{
int v;
if(c*amount>=sum/2)             //相当于物体对于容量为sum/2的背包,个数无限多,所以当成完全背包处理
{
cpl(c,w);
return;
}
v=1;
while(v<=amount)        //转化成0-1背包处理,用二进制压缩进行优化
{
one(c*v,w*v);
amount-=v;
v*=2;
}
one(c*amount,amount*w);
}
int main()
{
int a[10],i,j,k,num=0;
while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])!=EOF&&(a[1]!=0||a[2]!=0||a[3]!=0||a[4]!=0||a[5]!=0||a[6]!=0))
{
printf("Collection #%d:\n",++num);
memset(dp,0,sizeof(dp));
sum=(a[1]+a[2]*2+a[3]*3+a[4]*4+a[5]*5+a[6]*6);
if(sum%2!=0)
{
printf("Can't be divided.\n\n");
continue;
}
for(i=1;i<=6;i++)
{
if(a[i])
muty(i,i,a[i]);
}
if(dp[sum/2]==sum/2)
printf("Can be divided.\n");
else
printf("Can't be divided.\n");
printf("\n");
}}