hdu 1059 Dividing 多重背包

点击打开链接链接

Dividing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17544    Accepted Submission(s): 4912

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.

 

Source

Mid-Central European Regional Contest 1999

 

Recommend

JGShining

给出价值分别为1 2 3 4 5 6 的石头的个数 问能不能平分 

一个容量为sum/2的背包

#include<cstdio>
#include<cstring>
int maxn(int a,int b)
{
    return a>b?a:b;
}
int cnt[10];
int w[22222];
int dp[22222];
int main()
{
    int sum;
    int cas=1;
    int xx;
    while(scanf("%d",&cnt[0])!=EOF)
    {
        sum=cnt[0];
        for(int i=1;i<6;i++)
        {
            scanf("%d",&cnt[i]);
            sum+=cnt[i]*(i+1);
        }
        if(!sum)
            break;
        xx=0;
        for(int i=0;i<6;i++)
        {
            for(int j=1;j<=cnt[i];j*=2)
            {
                 w[xx++]=j*(i+1);
                 cnt[i]-=j;
            }
            if(cnt[i])
            {
                w[xx++]=cnt[i]*(i+1);
                cnt[i]=0;
            }
        }
        printf("Collection #%d:
",cas++);
        //printf("%d
",sum);
        if(sum%2)
        {
            printf("Can't be divided.
");
        }
        else
        {
            memset(dp,0,sizeof(dp));
            for(int i=0;i<xx;i++)
            {
                for(int j=sum/2;j>=w[i];j--)
                {
                    dp[j]=maxn(dp[j],dp[j-w[i]]+w[i]);
                }
            }
            //printf("%d
",dp[sum/2]);
            if(dp[sum/2]==sum/2)
            {
                printf("Can be divided.
");
            }
            else
            {
                printf("Can't be divided.
");
            }
        }
        puts("");
    }
    return 0;
}

版权声明：本文博客原创文章。博客，未经同意，不得转载。