poj 3414 Pots（Bfs）

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 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, andC. 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 operationsK. 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

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

題意：給出兩個容積分別爲 a 和 b的pot，按照以下三種操作方式，求出能否在一定步數後，使者兩個pot的其中一個的水量爲c。

     1.FILL(i)：將ipot倒滿水。

      2.DROP(i)：將ipot倒空水。

     3.POUR(i,j)： 將ipot的水倒到jpot上，直至要麼ipot爲空，要麼jpot爲滿。

 

直接BFS求解，將步驟存到結構體內即可：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <cstring>

using namespace std;

int A,B,C;

int flag[101][101];

char str[6][10]= {"FILL(1)","FILL(2)","POUR(2,1)","POUR(1,2)","DROP(1)","DROP(2)"};

struct node

{

    int num1,num2;

    string a;

    int step;

};

void bfs()

{

    node p,temp;

    int i;

    p.num1=0;

    p.num2=0;

    p.step=0;

    p.a="";

    queue<node>Q;

    while(!Q.empty())

        Q.pop();

    Q.push(p);

    while(!Q.empty())

    {

        p=Q.front();

        Q.pop();

        if(flag[p.num1][p.num2])

            continue;

        else

            flag[p.num1][p.num2]=1;

        if(p.num1==C||p.num2==C)

        {

            printf("%d\n",p.step);

            int j;

            for(j=0; j<p.step; j++)

            {

                printf("%s\n",str[(p.a[j])-'0']);

            }

            return;

        }

        for(i=0; i<4; i++)

        {

            temp=p;

            temp.step=p.step+1;

            if(i==0)

            {

                if(temp.num1==0)

                {

                    temp.num1=A;

                    temp.num2=p.num2;

                    temp.a=p.a+'0';

                    Q.push(temp);

                }

                if(temp.num2==0)

                {

                    temp.num2=B;

                    temp.num1=p.num1;

                    temp.a=p.a+'1';

                    Q.push(temp);

                }

            }

            if(i==1)

            {

                if(temp.num1!=0)

                {

                    temp.num1=0;

                    temp.num2=p.num2;

                    temp.a=p.a+'4';

                    Q.push(temp);

                }

                if(temp.num2!=0)

                {

                    temp.num2=0;

                    temp.num1=p.num1;

                    temp.a=p.a+'5';

                    Q.push(temp);

                }

            }

            if(i==2)

            {

                if(temp.num1!=0 && temp.num2!=B)

                {

                    int dd=B-temp.num2;

                    if(temp.num1>=dd)

                    {

                        temp.num2=B;

                        temp.num1-=dd;

                    }

                    else

                    {

                        temp.num2+=temp.num1;

                        temp.num1=0;

                    }

                    temp.a=p.a+'3';

                    Q.push(temp);

                }

            }

            if(i==3)

            {

                if(temp.num2!=0 && temp.num1!=A)

                {

                    int dd=A-temp.num1;

                    if(temp.num2>=dd)

                    {

                        temp.num1=A;

                        temp.num2-=dd;

                    }

                    else

                    {

                        temp.num1+=temp.num2;

                        temp.num2=0;

                    }

                    temp.a=p.a+'2';

                    Q.push(temp);

                }

            }

        }

    }

    printf("impossible\n");

}

int main()

{

    while(scanf("%d%d%d",&A,&B,&C)!=EOF)

    {

        memset(flag,0,sizeof(flag));

        bfs();

    }

    return 0;
}