POJ 3414--Pots（BFS+回溯路径）

Pots

Description

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

1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
2. DROP(i)      empty the pot i to the drain;
3. POUR(i,j)    pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i 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 exactly C 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)

和之前做过的一道二维bfs一样。仅仅只是这个须要回溯路径。非常easy在结构体中加一个变量来记录上一个状态在队列中的下标（手敲的队列比較好。这个时候在用STL队列好像不慷慨便）最后找到满足条件的状态逆向打印路径（由于记录的都是上一个状态）

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <queue>
using namespace std;
int m,n,c,s,e,p;
typedef struct node
{
	int v1,v2,cur,pre,op;
};
bool vis[999][999];
int ans[10010];
node que[10010];
void bfs()
{
	p=0;s=0;e=0;int pos;
	node t={0,0,0,0,0};
	que[e++]=t;
	vis[0][0]=1;
	while(s<e)
	{
		node f=que[s];pos=s;s++;
		if(f.v1==c||f.v2==c)
		{
			printf("%d\n",f.op);
			int tem=pos;
			for(int i=0;i<f.op;i++)
			{
               ans[p++]=que[tem].cur;
               tem=que[tem].pre;
			}
			for(int i=p-1;i>=0;i--)
			{
				switch(ans[i])
				{
					case 1:printf("FILL(1)\n");break;
					case 2:printf("FILL(2)\n");break;
					case 3:printf("DROP(1)\n");break;
					case 4:printf("DROP(2)\n");break;
					case 5:printf("POUR(2,1)\n");break;
					case 6:printf("POUR(1,2)\n");break;
				}
			}
			return ;
		}
		if(f.v1!=m)
		{
			t.v1=m;
			t.op=f.op+1;
			t.v2=f.v2;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 t.cur=1;
			 t.pre=pos;
			 que[e++]=t;
			}
		}
		if(f.v2!=n)
		{
			t.v2=n;
			t.op=f.op+1;
			t.v1=f.v1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 t.cur=2;
			t.pre=pos;
			 que[e++]=t;
			}
		}
		if(f.v1!=0)
		{
			t.v1=0;
			t.v2=f.v2;
			t.op=f.op+1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 t.cur=3;
			 t.pre=pos;
			 que[e++]=t;
			}
		}
		if(f.v2!=0)
		{
			t.v2=0;
			t.v1=f.v1;
			t.op=f.op+1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 t.cur=4;
			 t.pre=pos;
			 que[e++]=t;
			}
		}
		if(f.v2!=0&&f.v1!=m)
		{
			t.v2=f.v2-(m-f.v1);if(t.v2<0) t.v2=0;
			t.v1=f.v1+f.v2;  if(t.v1>m) t.v1=m;
			t.op=f.op+1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 t.cur=5;
			 t.pre=pos;
			 que[e++]=t;
			}
		}
		if(f.v1!=0&&f.v2!=n)
		{
			t.v1=f.v1-(n-f.v2);if(t.v1<0) t.v1=0;
			t.v2=f.v2+f.v1;  if(t.v2>n) t.v2=n;
			t.op=f.op+1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 t.cur=6;
			 t.pre=pos;
			 que[e++]=t;
			}
		}
	}
	puts("impossible");
}
int main()
{
	while(cin>>m>>n>>c)
	{
		memset(vis,0,sizeof(vis));
		bfs();
	}
	return 0;
}

版权声明：本文博客原创文章。博客，未经同意，不得转载。