Mother's Milk

Mother's Milk

Farmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk. Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.

Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.

PROGRAM NAME: milk3

INPUT FORMAT

A single line with the three integers A, B, and C.

SAMPLE INPUT (file milk3.in)

8 9 10

OUTPUT FORMAT

A single line with a sorted list of all the possible amounts of milk that can be in bucket C when bucket A is empty.

SAMPLE OUTPUT (file milk3.out)

1 2 8 9 10

SAMPLE INPUT (file milk3.in)

2 5 10

SAMPLE OUTPUT (file milk3.out)

5 6 7 8 9 10

肉流满面终于ac了！！！每次偷懒递归太多层就stackoverflow。。

/*
ID: des_jas1
PROG: milk3
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <string.h>
#include <algorithm>
#include <queue>
//#define fin cin
//#define fout cout
using namespace std;

int capacity[3],bb[3];
bool ans[25],k[21][21][21];


int main() 
{
	ofstream fout ("milk3.out");
   ifstream fin ("milk3.in");
	memset(ans,0,sizeof(ans));
	memset(bb,0,sizeof(bb));
	memset(k,0,sizeof(k));
	fin>>capacity[0]>>capacity[1]>>capacity[2];
	bb[2]=capacity[2];
    queue<int> p;
	p.push(0);
	p.push(0);
	p.push(bb[2]);
	int i,j,a,b;
  while(!p.empty())
  {	
	for(i=0;i<3;i++)
	{
		bb[i]=p.front();
	    p.pop();
	}
	if(!bb[0]) //若A是空的
		if(!ans[bb[2]]) //判断c此时是否有被记录过
			ans[bb[2]]=true;	
	bool used[3];
	memset(used,0,sizeof(used));
	for(i=0;i<3;i++)	
	{		
		if(bb[i]) //取某一个非空罐子		
		{			
			used[i]=true; //表示用它pour			
			for(j=0;j<3;j++)			
			{					
				if(!used[j]) //寻找另外两个罐子					
				{					
					if(bb[j]==capacity[j]) //被倒罐子满就跳过					
						break;					
					a=bb[j]; //把当前俩罐子的容量记录西来					
					b=bb[i];
					while(bb[j]<capacity[j] && bb[i]) //一直到被倒罐子满了 或者 另一个罐子空了					
					{					
						bb[j]++;					
						bb[i]--;						
					}	
					if(!k[bb[0]][bb[1]][bb[2]])
					{
						k[bb[0]][bb[1]][bb[2]]=true;
					    p.push(bb[0]);
					    p.push(bb[1]);
					    p.push(bb[2]);
					}
					bb[j]=a; //复原 					
					bb[i]=b;											
				}				
			}			
			used[i]=false;			
		}		
	}
  }
	for(i=0;ans[i]==false;i++);
	fout<<i;
	++i;
	for(;i<=20;i++)
		if(ans[i])
			fout<<" "<<i;
	fout<<endl;
	fout.close();
	fin.close();
    return 0;
}