POJ 3414 Pots BFS + 路径打印

第一次做路径打印的题，看了别人的博客才会，其基本思想就是在结构体内定义一个指针，指向生成该状态的状态，在达到目标状态后，用栈存储每一步的状态后输出即可，详情看代码：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
struct node
{
	int a , b , step , oper;
	node * pre;
}arr[400];
int A,B,C,ans,vis[105][105];
queue  q;
stack  stk;
void BFS()
{
	while(!q.empty())	q.pop();
	int cnt = -1;
	node now , next;
	now.a = now.b = now.step = 0;
	now.pre = NULL;
	q.push(now);
	vis[0][0] = 1;
	while(!q.empty())
	{
		now = q.front();
		cnt++;
		arr[cnt] = now;
		q.pop();
		for(int i = 1 ; i <= 6 ; i++)
		{
			switch(i)
			{
				case 1: next.a = A ; next.b = now.b; next.oper = 1 ; break;
				case 2: next.a = now.a ; next.b = B; next.oper = 2 ; break;
				case 3:	next.a = 0 ; next.b = now.b; next.oper = 3 ; break;
				case 4:	next.a = now.a ; next.b = 0; next.oper = 4 ; break;
				case 5:
					if(now.a > B - now.b)
					{
						next.a = now.a - B + now.b;
						next.b = B;
					}
					else
					{
						next.a = 0;
						next.b = now.b + now.a;
					}
					next.oper = 5;
					break;
				case 6:
					if(now.b > A - now.a)
					{
						next.a = A;
						next.b = now.b - A + now.a;
					}
					else
					{
						next.a = now.b + now.a;
						next.b = 0;
					}
					next.oper = 6;
					break;
			}
			if(vis[next.a][next.b])	continue;
			vis[next.a][next.b] = 1;
			next.step = now.step + 1;
			next.pre = &arr[cnt];    //记录每个状态的父状态
			if(next.a == C || next.b == C)
			{
				ans = next.step;
				while(!stk.empty())	stk.pop();
				while(next.pre)
				{
					stk.push(next.oper); //将每一步的操作压入栈
					next = *next.pre;    		
		                }
				return ;
			}
			q.push(next);
		}
	}
	return ;
}
int main()
{
	while(cin>>A>>B>>C)
	{
		memset(vis,0,sizeof(vis));
		ans = 0 ;
		BFS();
		if(!ans)	cout<<"impossible"<