BFS很簡單的思想，但是注意剪枝，因為很多會重復，比如，不斷的empty，這個重復很嚴重，所以很有必要去除重復，即記錄1000 *1000的矩陣，保證對想通的a,b不重復計算
#include
#include
#include
#include
using namespace std;
string op[6]={fill A,fill B,pour B A,pour A B,empty A,empty B};
bool visited[1001][1001];
struct Node
{
int left,right;
vector op;
};
void Cal(int a,int b,int q)
{
queue result;
Node first,second;
first.left=0;
first.right=0;
result.push(first);
memset(visited,0,sizeof(visited));
while(!result.empty())
{
first=result.front();
result.pop();
for(int i=0;i<6;i++)
{
second=first;
switch (i)
{
case 0:
second.left=a;
break;
case 1:
second.right=b;
break;
case 2://pour b to a
if(second.right<=a-second.left)
{
second.left=second.left+second.right;
second.right=0;
}
else
{
second.right=second.right-(a-second.left);
second.left=a;
}
break;
case 3:
if(second.left<=b-second.right)
{
second.right=second.right+second.left;
second.left=0;
}
else
{
second.left=second.left-(b-second.right);
second.right=b;
}
break;
case 4:
second.left=0;
break;
case 5:
second.right=0;
break;
}
second.op.push_back(i);
if(second.right==q)
{
for(int i=0;i
