POJ3414 Pots【BFS】

Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23177		Accepted: 9827		Special Judge

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 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)

Source

Northeastern Europe 2002, Western Subregion

题目链接：POJ3414 Pots

问题描述：给你两个空壶，容量分别为A,B。定义3中操作，FILL（i）表示把壶i装满，DROP（i）表示把壶清空，POUR（i,j）表示把壶i中的水倒入j中，直到i变空，或者j变满。问经过若干步操作后，是否能使得其中一个壶的容量为C，如果可以则打印步数和相应步骤，否则输出“impossible”

题解：初始状态为把壶a装满或者把壶b装满，然后对每个状态进行6中操作：装满壶a，转满壶b，清空壶a，清空壶b，壶a倒入壶b，壶b倒入壶a，详情如下

AC的C++程序：

#include<iostream>
#include<queue>
#include<string>
#include<cstring>
#define MIN(x,y) x<y?x:y
using namespace std;
const int N=500;
struct Node{
	int a,b,pre,id;
	string s;
}st[2*N];

int A,B,C;
bool vis[N][N];

int bfs()
{
	memset(vis,false,sizeof(vis));
	queue<Node>q;
	//初始有两个状态，要么把a装满，要么把b装满 
	st[0].a=A,st[0].b=0,st[0].pre=-1,st[0].id=0,st[0].s="FILL(1)";
	st[1].a=0,st[1].b=B,st[1].pre=-1,st[1].id=1,st[1].s="FILL(2)";
	q.push(st[0]);
	q.push(st[1]);
	vis[A][0]=vis[0][B]=true;
	int k=2;
	while(!q.empty()){
		Node f=q.front();
		q.pop();
		if(f.a==C||f.b==C)//如果转换成功，返回此状态的编号 
		  return f.id;
		//FILL操作
		Node e=f;
		if(e.a!=A&&!vis[A][e.b]){
			e.a=A,e.id=k,e.pre=f.id,e.s="FILL(1)";
			st[k++]=e; 
			vis[A][e.b]=true;
			q.push(e);
		}
		e=f;
		if(e.b!=B&&!vis[e.a][B]){
			e.b=B,e.id=k,e.pre=f.id,e.s="FILL(2)";
			st[k++]=e; 
			vis[e.a][B]=true;
			q.push(e);
		}
		//DROP操作
		e=f;
		if(e.a!=0&&!vis[0][e.b]){
			e.a=0,e.id=k,e.pre=f.id,e.s="DROP(1)";
			st[k++]=e; 
			vis[0][e.b]=true;
			q.push(e);
		}
		e=f;
		if(e.b!=0&&!vis[e.a][0]){
			e.b=0,e.id=k,e.pre=f.id,e.s="DROP(2)";
			st[k++]=e; 
			vis[e.a][0]=true;
			q.push(e);
		}
		//POUR操作
		e=f;
		if(e.a!=0){
			int d=MIN(e.a,B-e.b);//a向b中倒水的最大量 
			if(!vis[e.a-d][e.b+d]){
				vis[e.a-d][e.b+d]=true;
				e.a-=d,e.b+=d,e.pre=f.id,e.id=k,e.s="POUR(1,2)";
				st[k++]=e;
				q.push(e);
			}
		}
		e=f;
		if(e.b!=0){
			int d=MIN(A-e.a,e.b);//b向a中倒水的最大量 
			if(!vis[e.a+d][e.b-d]){
				vis[e.a+d][e.b-d]=true;
				e.a+=d,e.b-=d,e.pre=f.id,e.id=k,e.s="POUR(2,1)";
				st[k++]=e;
				q.push(e);
			}
		}
	}
	return -1; //表示不能成功 
}
void print(int i)
{
	if(st[i].pre!=-1)
	  print(st[i].pre);
	cout<<st[i].s<<endl;
}

int main()
{
	scanf("%d%d%d",&A,&B,&C);
	if(C==0){
		printf("0\n");
		return 0;
	}
	int id=bfs();//终止状态id
	if(id==-1){
		printf("impossible\n");
		return 0;
	}
	int cnt=1,i=id;
	while(st[i].pre!=-1){
		i=st[i].pre;
		cnt++;
	}
	printf("%d\n",cnt);
	print(id);
	return 0;
}

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