pots 倒水问题(bfs)

最新推荐文章于 2021-12-10 17:39:55 发布

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本文介绍了一个经典的水罐问题，通过广度优先搜索(BFS)算法找到从两个不同容量的水罐中得到指定水量的最短操作序列。讨论了如何避免状态重复，并详细展示了实现过程。

点击打开链接 <----HDU

Problem Description

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
DROP(i) empty the pot i to the drain;
POUR(i,j) pour from pot i to pot j; after this operation either the potj is full (and there may be some water left in the pot i), or the poti is empty (and all its contents have been moved to the pot j).

Write a program to find the shortest possible sequence of these operations that will yield exactlyC liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input

  
   3 5 4

Sample Output

  
   6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

//广搜 打印路径问题  
//特别新的就是广搜打印出路径来了 
//主要是怎么走理清思路，不要重复。 
#if 0
#include<iostream>
#include<queue>

using namespace std;
struct go
{
	int  x,y,t;
	string s;
};

int m,n,c;
bool vis[105][105];
bool flag=0;
void sprint(string s)
{
	for(int i=0; i<s.size(); i++)
	{
		if(s[i]=='1')
			cout<<"FILL(1)"<<endl;
		else
		if(s[i]=='2')
			cout<<"FILL(2)"<<endl;
		else
		if(s[i]=='3')
			cout<<"POUR(2,1)"<<endl;
		else
		if(s[i]=='4')
			cout<<"POUR(1,2)"<<endl;
		else
		if(s[i]=='5')
			cout<<"DROP(1)"<<endl;
		else
		if(s[i]=='6')
			cout<<"DROP(2)"<<endl;					
	}
	
}


void bfs()
{
	queue<go> q;
	go p;
	p.t=1;
	p.x=m;
	p.y=0;
	p.s='1';
	q.push(p);
	p.t=1;
	p.x=0;
	p.y=n;
	p.s='2';
	q.push(p);
	
	while(!q.empty())
	{
		go now;
		now=q.front();
		q.pop(); 
		
		if(now.x==c||now.y==c)
		{
			cout<<now.t<<endl;
			sprint(now.s);
			flag=1;
			return;
		}
		
		if(now.x<m)                //自来水加水 
		{
			go a;
			a.x=m;
			a.y=now.y;
			
			if(vis[a.x][a.y]==0)
			{
				a.t=now.t+1;
				vis[a.x][a.y]=1;
				a.s=now.s+'1';
				q.push(a);
			}
			
			if(now.y>0)         //水桶加水 
			{
				int cx=m-now.x;
				if(now.y>=cx)
				{
					a.x=m;
					a.y=now.y-cx;
				}
				else
				{
					a.x=now.x+now.y;
					a.y=0;
				}
				if(vis[a.x][a.y]==0)
				{
					a.t=now.t+1;
					vis[a.x][a.y]=1;
					a.s=now.s+'3';
					q.push(a);
				}
			}
		}
		
		if(now.y<n)       //自来水加水 
		{
			go a;
			a.x=now.x;
			a.y=n;
			
			if(vis[a.x][a.y]==0)
			{
				a.t=now.t+1;
				vis[a.x][a.y]=1;
				a.s=now.s+'2';
				q.push(a);
			}
			
			if(now.x>0)    //水桶加水 
			{
				int cy=n-now.y;
				if(now.x>=cy)
				{
					a.y=n;
					a.x=now.x-cy;
				}
				else
				{
					a.y=now.x+now.y;
					a.x=0;	
				}
				if(vis[a.x][a.y]==0)
				{
					a.t=now.t+1;
					vis[a.x][a.y]=1;
					a.s=now.s+'4';
					q.push(a);	
				}	
			}	
		}
		
		if(now.x>0)     //倒水 
		{
			go a;
			a.x=0;
			a.y=now.y;	
			if(vis[a.x][a.y]==0)
			{
				a.t=now.t+1;
				vis[a.x][a.y]=1;
				a.s=now.s+'5';
				q.push(a);	
			}	
		}
		if(now.y>0)  //倒水 
		{
			go a;
			a.y=0;
			a.x=now.x;
			if(vis[a.x][a.y]==0)
			{
				a.t=now.t+1;
				vis[a.x][a.y]=1;
				a.s=now.s+'6';
				q.push(a);	
			}		
		}		
	}
}

int main()
{
	cin>>m>>n>>c;
	vis[0][0]=1;
	bfs();
	if(flag==0)
	{
		cout<<"impossible"<<endl;
	}
}
#endif