hdu-1059-Dividing（多重背包+二进制优化）

Dividing

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27731 Accepted Submission(s): 7966

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：
输入一行 6个数a[1] ~ a[6]，下标为价值 ，数为这种价值的有多少个（输入全为0时结束输入）。求是否能够本平分
思路: 多重背包，+二进制优化。

code

#include
#include
#include
#define inf 0x3f3f3f3f
using namespace std;
int dp[202000],v[100],num[100];
int sum ;
//m背包的总容量、v物品的体积、w物品的价值
void OneZeroPack(int m,int v,int w)  //0-1背包
{
    for(int i=m;i>=v;i--)
        dp[i]=max(dp[i],dp[i-v]+w);
}

//m背包的总容量、v物品的体积、w物品的价值
void CompletePack(int m,int v,int w)  //完全背包
{
    for(int i=v;i<=m;i++)
        dp[i]=max(dp[i],dp[i-v]+w);
}

//m背包的总容量、v物品的体积、w物品的价值、num物品的数量
void MultiplePack(int m,int v,int w,int num)//多重背包
{
    if(v*num>=m)
    {
        CompletePack(m,v,w);
        return ;
    }
    int k=1;
    for(k=1;k<=num;k<<=1)
    {
        OneZeroPack(m,k*v,k*w);
        num=num-k;
    }
    if(num)
        OneZeroPack(m,num*v,num*w);
}
int main()
{
    int falg=0;
    while(cin>>num[1]>>num[2]>>num[3]>>num[4]>>num[5]>>num[6]){
        if(!(num[1]||num[2]||num[3]||num[4]||num[5]||num[6])) break;
        printf("Collection #%d:\n",++falg);
        sum =num[1]+2*num[2]+3*num[3]+4*num[4]+5*num[5]+6*num[6];
        if (sum % 2 != 0) ///当sum 不能被整除时 肯定是不能平分的所以 直接输出
        {
            printf("Can't be divided.\n\n");
            continue;
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=6;i++){
            MultiplePack(sum/2,i,i,num[i]);
        }
       // cout<