Mother's Milk

http://train.usaco.org/usacoprob2?a=EN6XVH05xAs&S=milk3

题目大意：

A，B，C三个容器，每个容器都有固定的体积，刚开始A，B是空的，C装满牛奶；三个容器中的牛奶可以相互倒，每次倒，直到一个容器倒空或另一个容器倒满为止；找出当A为空是，C容器中可能剩下的牛奶；

#include <iostream>
#include <fstream>
#include <algorithm>
//倒水的策略：
using namespace std;
struct bucket
{
	int capacity;
	int  state; 
} a[3];
bool ans[22], f[22][22][22];		//f[a][b][c]:判断这种状态是否搜索过 
void pour(bucket& to, bucket& from)
{
	if(to.capacity - to.state > from.state) to.state += from.state, from.state = 0;
	else from.state = from.state - (to.capacity - to.state), to.state = to.capacity;
}
void dfs()
{
	if(f[a[0].state][a[1].state][a[2].state]) return;
	if(a[0].state == 0)
		if(!ans[a[2].state]) {ans[a[2].state] = true;}
	f[a[0].state][a[1].state][a[2].state] = true;
	bucket t0 = a[0], t1 = a[1], t2 = a[2];
	for(int i = 0; i < 3; i++)
	{
		for(int j = 0; j < 3; j++)
		{
			if(i == j) continue;
			pour(a[j], a[i]);
			dfs();
			a[0] = t0;
			a[1] = t1;
			a[2] = t2;
		}
	}
}
int main()
{
	ifstream fin("milk3.in");
	ofstream fout("milk3.out");
	fin >> a[0].capacity >> a[1].capacity >> a[2].capacity;
	a[0].state = a[1].state = 0;
	a[2].state = a[2].capacity;
	dfs();
	for(int i = 0; i <= a[2].capacity; i++)
	{
		if(i > 0 && ans[i - 1]) fout <<" ";
		if(ans[i]) fout << i;
	}
	fout << endl;
	fout.close();
	return 0;
}