POJ1014解题报告

Dividing

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43674		Accepted: 10841

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.

题意很简单，就是有一组大理石，即每一行输入。行的列号位大理石的价值，值为大理石的数目。如1 0 1 2 0 0即价值为1的大理石有1个，价值为2的有0个，价值为3的有1个，以此类推。判断是否有一种分法，使得Marsha和Bill得到的大理石价值相等，即平分大理石。

也是搜索的题目，首先如果大理石价值本身不为偶数，当然不能平分；进入搜索，剪掉所有组合值会大于总价值一半的枝，当搜索到的组合恰好等于总价值的一半时，搜索成功！

与1010思路差不多。

被一位同学指出了问题，说 0 0 0 6 2 2不能给出正确的切分答案。看了一下，确实是回溯的时候有点问题。改后的代码如下：

using namespace std;
int marble[6];
int found;
int targetNum,i;
void calculate(int currentNum,int i)
{
    if(currentNum*2==targetNum)
    {
         found=1; 
         return;
    }
    for(int k=i;k>=0;k--)
    {
         if(marble[k])
         {
              if(k+1+currentNum<=targetNum/2)
              {
                 marble[k]--;
                 calculate((k+1)+currentNum,k);
                 if(found)
                 {
                     break; 
                 }
                 marble[k]++;
              }
         }
    }
}
void init()
{
    int index=0;
    while(cin>>marble[0]>>marble[1]>>marble[2]>>marble[3]>>marble[4]>>marble[5])
    {
        found=0;
        targetNum=0;
        for(i=0;i<6;i++)
        {
            targetNum+=(i+1)*marble[i]; 
        } 
     
        if(targetNum==0)
        {
            break; 
        }
        
        if(targetNum%2!=0)
        {
             found=0;
        }
        else
        {
             calculate(0,5);
        }
        if(found)
        {
            cout<<"Collection #"<<++index<<":"<

与之前的答案仅有一行不同。即：
marble[k]--;
calculate((k+1)+currentNum,k);
if(found)
{
 break;
 }
 marble[k]++;
因为calculate没得到解退出后，回溯时需要将marble数组的值还原。

PS:当前的POJ仿佛对时间的要求更加严格了，故虽然解答是正确的，但真的要直接粘贴提交的话会超时。。。。反正我当时是AC了的。