POJ 3414 Pots 记录路径的广搜

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)

题意：有两个空瓶 a，b是它们的容量，c是容量目标。 能够有三种操作 充满随意一瓶。倒空随意一瓶，将随意一瓶倒入还有一瓶（能剩下但不能溢出）；求随意一瓶的体积
达到目标体积所须要的最小操作数。并依此输出该操作。
代码：
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
int m,n,k;
int vis[105][105];
char opr[20][20]= {" " , "FILL(1)" , "FILL(2)" , "DROP(1)" , "DROP(2)" , "POUR(1,2)" , "POUR(2,1)" };     //共6种操作
struct node
{
    int x,y,step;
    int w[200];                    //用来记录路径的数组数组  
};
void bfs( )
{
    int i,j;
    int kx,ky;
    memset(vis,0,sizeof(vis));
    queue<node>q;
    node now,next;
    now.x=0,now.y=0,now.step=0;
    q.push(now);
    vis[0][0]=1;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(now.x==k||now.y==k)
        {
            cout<<now.step<<endl;
            for( i=1; i<=now.step; i++)
            {
                cout<<opr[now.w[i]]<<endl;
            }
            return;
        }
        for(i=0; i<6; i++)        // 共六种操作
        {
            if(i==0)
            {
                next.y=now.y;
                next.x=m;
                now.w[now.step+1]=1;           //随时更新
            }
            else if(i==1)
            {
                next.x=now.x;
                next.y=n;
                now.w[now.step+1]=2;
            }
            else if(i==2)
            {
                next.y=now.y;
                next.x=0;
                now.w[now.step+1]=3;
            }
            else if(i==3)
            {
                next.x=now.x;
                next.y=0;
                now.w[now.step+1]=4;
            }
            else if(i==4)
            {
                if(n-now.y>now.x)          //每种pour 应有两种情况
                {
                    next.x=0;
                    next.y=now.x+now.y;
                }
                else
                {
                    next.y=n;
                    next.x=now.x-n+now.y;
                }
                now.w[now.step+1]=5;
            }
            else if(i==5)
            {
                if(m-now.x>now.y)
                {
                    next.y=0;
                    next.x=now.x+now.y;
                }
                else
                {
                    next.x=m;
                    next.y=now.y-m+now.x;
                }
                now.w[now.step+1]=6;             
            }
            if(vis[next.x][next.y]==1)
                continue;
            vis[next.x][next.y]=1;
            next.step=now.step+1;
            for(j=1; j<=next.step; j++)            //记录之前的行动  
            {
                next.w[j]=now.w[j];
            }
            q.push(next);
        }
    }
    cout<<"impossible"<<endl;             //不要忘了这个情况
    return;
}
int main()
{
    int i;
    while(cin>>m>>n>>k)
    {
        bfs();
    }
    return 0;
}

通过自己对队列的理解成功写出了代码 还是挺开心的。
。。