POJ 1014 / HDU 1059 Dividing 多重背包+二进制分解

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：给出6种大小的石头，问能不能分成数值一样的两堆。

思路：背包方面是简单的，但主要的问题是，如果一个一个枚举物品，物品太多肯定会超时。所以就有将物品的数量进行二进制分解，即分成2的幂次，每个2的幂都作为一个新的物品，再进行01背包就行了。POJ上交不但不支持<bits/stdc++.h>而且数组还要开大一点/.\

/** @Date    : 2016-12-10-20.39
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version :
  */
#include
#include
#include
#include
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

int cnt = 0;
int a[10];
int b[N];

int dp[60010];

void binaDiv(int x, int v)
{
    int r = 1;
    while(r < x)
    {
        b[cnt++] = r * v;
        x -= r;
        r <<= 1;
    }
    b[cnt++] = x * v;
}

int main()
{
    int c = 0;
    while(true)
    {
        int sum = 0;
        for(int i = 1; i <= 6; i++)
            scanf("%d", a + i), sum += a[i]*i;
        if(sum == 0)
            break;
        printf("Collection #%d:\n", ++c);
        if(sum & 1)
        {
            printf("Can't be divided.\n\n");
            continue;
        }
        sum /= 2;

        cnt = 0;
        for(int i = 1; i <= 6; i++)
        {
            binaDiv(a[i], i);
        }
        MMF(dp);
        for(int i = 0; i < cnt; i++)
        {
            for(int j = sum; j >= 0; j--)
            {
                if(j >= b[i])
                    dp[j] = max(dp[j], dp[j - b[i]] + b[i]);
            }
        }
        if(dp[sum] == sum)
            printf("Can be divided.\n\n");
        else printf("Can't be divided.\n\n");
    }
    return 0;
}