poj 1014 Dividing 背包问题(dfs)(其实我用最笨的方法写的)

本文介绍了一个关于公平分配问题的解决方案,使用C++编程语言实现。该问题涉及将不同价值的物品(如大理石)分为两组,使每组的总价值相等。通过迭代和条件判断的方法来检查是否存在可行的分配方案。
 不知道怎么回事,还是没有ac。
dfs写出来一直没反映,于是就这么写了。dfs还需要练习啊。
艹!

Dividing
Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49516		Accepted: 12486
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 file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , 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 collection, 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.
#include<iostream>
using namespace std;
int beibao[7];
int test=0;
int halfvalue=0;
bool flag=false;
int value;
int main()
{

    while(cin>>beibao[1]>>beibao[2]>>beibao[3]>>beibao[4]>>beibao[5]>>beibao[6])
    {
        flag=false;
        value=0;
        halfvalue=0;
        if(beibao[1]==0&&beibao[2]==0&&beibao[3]==0&&beibao[4]==0&&beibao[5]==0&&beibao[6]==0) break;
        for(int i=1;i<7;++i)
            halfvalue=i*beibao[i]+halfvalue;
        halfvalue=halfvalue/2;

        if(halfvalue%2)
        {
            cout<<"Collection #"<<++test<<':'<<endl;
            cout<<"Can't be divided."<<endl<<endl;
            continue;
        }


       int i=1;
       while(i<=6)
              {
                             while(beibao[i])
                              {
                               beibao[i]--;
                               value=value+i;
                               if(value<halfvalue)  continue;
                               if(value==halfvalue)  {flag=true;break;}
                               if(value>halfvalue)   break;
                               }
                            i++;
                            if(value<halfvalue)  continue;
                            if(value==halfvalue)  {flag=true;break;}
                            if(value>halfvalue)   break;

                }
      if(flag)
        {
            cout<<"Collection #"<<++test<<':'<<endl;
            cout<<"Can be divided."<<endl<<endl;
            continue;
        }
        else
        {
            cout<<"Collection #"<<++test<<':'<<endl;
            cout<<"Can't be divided."<<endl<<endl;
            continue;
        }
    }
}

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