HDU1059 && POJ1014 :Dividing(多重背包)

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0

 

[align=left]Sample Output[/align]

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

 

 

题意：一共有6种物品，价值分别是1,2,3,4,5,6,给出每种物品的数目，判断师傅能够平分

思路：典型的多重背包，防止超时要进行一些必要的剪枝和二进制优化

 

[cpp] view
plain copy

#include <stdio.h>  

#include <algorithm>  

#include <string.h>  

using namespace std;  

  

int dp[100000];  

  

int main()  

{  

    int a[11],sum,v,i,j,k,cnt,cas = 1;  

    while(~scanf("%d",&a[1]))  

    {  

        sum = a[1];  

        for(i = 2;i<=6;i++)  

        {  

            scanf("%d",&a[i]);  

            sum+=i*a[i];  

        }  

        if(!sum)  

        break;  

        printf("Collection #%d:\n",cas++);  

        if(sum%2)//总和为奇数，必定不能平分  

        {  

            printf("Can't be divided.\n\n");  

            continue;  

        }  

        v = sum/2;  

        memset(dp,0,sizeof(dp));  

        dp[0] = 1;  

        for(i = 1;i<=6;i++)  

        {  

            if(!a[i])  

            continue;  

            for(j = 1;j<=a[i];j*=2)//二进制优化  

            {  

                cnt = j*i;  

                for(k = v;k>=cnt;k--)  

                {  

                    if(dp[k-cnt])//必须前面的能够放入背包，现在的才能放入背包  

                    dp[k] = 1;  

                }  

                a[i]-=j;  

            }  

            cnt = a[i]*i;//剩下的  

            if(cnt)  

            {  

                for(k = v;k>=cnt;k--)  

                {  

                    if(dp[k-cnt])  

                    dp[k] = 1;  

                }  

            }  

        }  

        if(dp[v])  

        printf("Can be divided.\n\n");  

        else  

        printf("Can't be divided.\n\n");  

    }  

  

    return 0;  

}