hdu 1059 Dividing(dp+二进制优化)

最新推荐文章于 2022-08-16 21:45:30 发布

原创最新推荐文章于 2022-08-16 21:45:30 发布 · 658 阅读

0 ·

CC 4.0 BY-SA版权

DP 专栏收录该内容

107 篇文章

订阅专栏

解决Marsha和Bill如何公平地分配不同价值的大理石集合的问题。通过二进制优化的背包算法检查是否存在一种分配方式使得两人的总价值相等。

Dividing

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10438 Accepted Submission(s): 2918

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

思路：直接背包的时间是6*N*N毫无疑问TLE。所以需要二进制优化。

        for(int i=1;i<=6;i++)
        for(int j=1;j<=v[i];j++)
        for(int k=sum-i*j;k>=0;k--)
        if(dp[k])dp[k+j*i]=1;

但仔细考虑这个过程，手动模拟一下你会发现很多的重复标记。

因此可以使用二进制来减小这些重复标记的过程。1 2, 4 8 16...

具体手动算一下，就明白了。

for(int i=1;i<=6;i++)
    { for(int j=1;j<=v[i];j*=2)
      { v[i]-=j;
        for(int k=sum-i*j;k>=0;k--)
        {
        if(dp[k])dp[k+j*i]=1;
        }
     }
     if(v[i])
     {
       for(int k=sum-i*v[i];k>=0;k--)
       if(dp[k])dp[k+i*v[i]]=1;
     }
    }

#include<iostream>
#include<cstring>
using namespace std;
const int mm=620010;
int dp[mm];
int v[7];
int main()
{ int sum;
  int cas=0;
  while(1)
  { sum=0;++cas;
    for(int i=1;i<=6;i++)
    {
      cin>>v[i];
      sum+=v[i]*i;
    }
    if(!sum)break;
    cout<<"Collection #"<<cas<<":\n";
    if(sum&1)
    {cout<<"Can't be divided.\n\n";
     continue;
    }
    sum/=2;
    memset(dp,0,sizeof(dp));
    dp[0]=1;
    for(int i=1;i<=6;i++)
    { for(int j=1;j<=v[i];j*=2)
      { v[i]-=j;
        for(int k=sum-i*j;k>=0;k--)
        {
        if(dp[k])dp[k+j*i]=1;
        }
     }
     if(v[i])
     {
       for(int k=sum-i*v[i];k>=0;k--)
       if(dp[k])dp[k+i*v[i]]=1;
     }
    }
    if(dp[sum])cout<<"Can be divided.\n";
    else cout<<"Can't be divided.\n";
    cout<<"\n";
  }
}