【原】 POJ 3414 Pots 状态BFS 解题报告

http://poj.org/problem?id=3414

方法：
该题简化之后即为求从初始状态(0,0)到终止状态(i,C)或(C,j)的最短路径
对于每个状态(i,j)存在由6种操作得到的6个邻接状态，即为图中的邻接节点
将每种操作和得到的节点状态对应上以便打印路径，即pathArr[6]和adjVertex[6]

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)

#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>

using namespace std ;

const int INF = 0x7fffffff ;

struct gSlot
{
    gSlot(){}
    gSlot(int i,int j):a(i),b(j){}
    int a ;
    int b ;
};

struct tSlot
{
    tSlot():dist(INF){preVertex.a=0 ; preVertex.b=0;}
    int dist ;
    gSlot preVertex ;
    int path ;  //结合pathArr的下标记录路径的操作
};

typedef vector< struct gSlot > Graph[101][101] ;
typedef tSlot Table[101][101] ;

Graph G ;
Table T ;
gSlot myQueue[60000] ;
string pathArr[6] = { "FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)" } ;

void PrintPath(int a,int b)
{
    if( a==0 && b==0 )
        return ;
    PrintPath( T[a][b].preVertex.a , T[a][b].preVertex.b ) ;
    cout<<pathArr[T[a][b].path]<<endl;
}

void run3414()
{
    int A,B,C ;
    int i,j ;
    int adji,adjj ;
    int k ;
    int front,rear,size ;  //myQueue
    int curDist ;
    gSlot adjVertex[6] ;  //通过计算得到在某状态时由6种操作得到的6个邻接状态

    scanf( "%d%d%d", &A,&B,&C );

    gSlot s(0,0) ;  //起点

    size = 0 ;      //queue初始化
    rear = 0 ;
    front = 1 ;

    T[0][0].dist = 0 ;    //table初始化
    myQueue[++rear] = s ; //起点入队
    ++size ;

    while( size != 0 )
    {
        i = myQueue[front].a ;   //得到出队节点和其距离
        j = myQueue[front].b ;
        curDist = T[i][j].dist ;
        ++front ;
        --size ;

        if( i==C || j==C )
        {
            printf( "%d\n", curDist );
            PrintPath(i,j) ;
            return ;
        }

        //得到(i,j)邻接节点，与pathArr的操作对应
        adjVertex[0].a = A; adjVertex[0].b = j;  //FILL(1) --> (A,j)
        adjVertex[1].a = i; adjVertex[1].b = B;  //FILL(2) --> (i,B)
        adjVertex[2].a = 0; adjVertex[2].b = j;  //DROP(1) --> (0,j)
        adjVertex[3].a = i; adjVertex[3].b = 0;  //DROP(2) --> (i,0)

        if( i+j>=B )  //POUR(1,2) --> (i+j-B,B) OR (0,i+j)
        {
            adjVertex[4].a = i+j-B; adjVertex[4].b = B;
        }
        else
        {
            adjVertex[4].a = 0; adjVertex[4].b = i+j;
        }

        if( i+j>=A )  //POUR(2,1) --> (A,i+j-A) OR (i+j,0)
        {
            adjVertex[5].a = A; adjVertex[5].b = i+j-A;
        }
        else
        {
            adjVertex[5].a = i+j; adjVertex[5].b = 0;
        }

        for( k=0 ; k<6 ; ++k )  //对邻接的6个节点依次处理
        {
            adji = adjVertex[k].a ;
            adjj = adjVertex[k].b ;

            if( T[adji][adjj].dist == INF )
            {
                T[adji][adjj].dist = curDist+1 ;
                T[adji][adjj].path = k ;   //对应pathArr中的操作
                T[adji][adjj].preVertex.a = i ;
                T[adji][adjj].preVertex.b = j ;

                //得到结果
                if( adji==C || adjj==C )
                {
                    printf( "%d\n", curDist+1 );
                    PrintPath(adji,adjj) ;
                    return ;
                }

                //入队
                ++size ;
                ++rear ;
                myQueue[rear].a = adji ;
                myQueue[rear].b = adjj ;
            }
        }
    }
    printf("impossible\n") ;
}