POJ 1014 Dividing (多重揹包)

Dividing

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 58921		Accepted: 15200

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

Mid-Central European Regional Contest 1999

题目大意：

有六个大理石，他们的价值分别是1,2,3,4,5,6，然后分别给出六个大理石的个数，问如何平分给两个人，令两个人所得到的价值相等。

这里要求恰好达到平分这个状态，也就是当揹包大小是价值的一半恰好装满，所以dp的初始值是负无穷。
开始以为数据小(没看清题)，其实大的该死，我就用01揹包去做，果断超时。
所以就改成多重揹包改二进制优化的01揹包+完全揹包。就是个模板了。好像只要优化的01揹包也能过，不过不知道怎么做。
 下面是模拟多重揹包问题，总共6个物品，将第i个物品的价值与费用都看做是i，这样套用揹包九讲的多重揹包模板就可以了，不过注意的是数组要开的大一点.
代码：16MS (这是codeblocks运行的代码)

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
#define M 100005
#define INF -99999;
int sum,v[7],dp[M];
void CompletePack(int cost,int wight)
{
    int i;
    for(i=cost;i<=sum;i++)
        dp[i]=max(dp[i],dp[i-cost]+wight);
}
void ZeroOnePack(int cost,int wight)
{
    int i;
    for(i=sum;i>=cost;i--)
        dp[i]=max(dp[i],dp[i-cost]+wight);
}
void MultiplePack(int cost,int wight,int amount)
{
    if(cost*amount>=sum)
    {
        CompletePack(cost,wight);
        return;
    }
    int k=1;
    while(k<amount)
    {
        ZeroOnePack(k*cost,k*wight);
        amount-=k;
        k=k<<1;
    }
    ZeroOnePack(amount*cost,amount*wight);
}
void DP()
{
    int i,j,k;
    for(i=0;i<=sum;i++) dp[i]=i==0?0:INF;
    for(i=1;i<=6;i++)
        MultiplePack(i,i,v[i]);
}
int main()
{
    int flag,i,t=0;
    while(1)
    {   t++;sum=0;
        for(i=1;i<=6;i++)
    {
        cin>>v[i];
        sum+=i*v[i];
    }
    if(!sum) break;
    printf("Collection #%d:\n",t);
        if(sum&1){
            printf("Can't be divided.\n\n");
            continue;
        }
        sum=sum>>1;
        DP();
        if(dp[sum]>=0)
            printf("Can be divided.\n\n");
        else
            printf("Can't be divided.\n\n");
    }
    return 0;
}