【原】 POJ 1606 Jugs 状态BFS 解题报告

 

http://poj.org/problem?id=1606

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)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]

注意：
此题的input无终止条件，所以一定要在scanf后面加上!=EOF，不然会造成Output Limit Exceeded

Description

In the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fill the 5-gallon jug with exactly 4 gallons. This problem generalizes that puzzle. 
You have two jugs, A and B, and an infinite supply of water. There are three types of actions that you can use: (1) you can fill a jug, (2) you can empty a jug, and (3) you can pour from one jug to the other. Pouring from one jug to the other stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons and B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A. 
A problem is given by a triple (Ca,Cb,N), where Ca and Cb are the capacities of the jugs A and B, respectively, and N is the goal. A solution is a sequence of steps that leaves exactly N gallons in jug B. The possible steps are 
fill A 
fill B 
empty A 
empty B 
pour A B 
pour B A 
success 
where "pour A B" means "pour the contents of jug A into jug B", and "success" means that the goal has been accomplished. 
You may assume that the input you are given does have a solution.

Input

Input to your program consists of a series of input lines each defining one puzzle. Input for each puzzle is a single line of three positive integers: Ca, Cb, and N. Ca and Cb are the capacities of jugs A and B, and N is the goal. You can assume 0 < Ca <= Cb and N <= Cb <=1000 and that A and B are relatively prime to one another.

Output

Output from your program will consist of a series of instructions from the list of the potential output lines which will result in either of the jugs containing exactly N gallons of water. The last line of output for each puzzle should be the line "success". Output lines start in column 1 and there should be no empty lines nor any trailing spaces.

Sample Input

3 5 4

5 7 3

Sample Output

fill B

pour B A

empty A

pour B A

fill B

pour B A

success

fill A

pour A B

fill A

pour A B

empty B

pour A B

success

 

#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] ;

gSlot myQueue[60000] ;
string pathArr[6] = { "fill A","fill B","empty A","empty B","pour A B","pour B A" } ;

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

void run1606()
{
    int A,B,C ;
    int i,j ;
    int adji,adjj ;
    int k ;
    int front,rear,size ;  //myQueue
    int curDist ;
    bool found ;
    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

    while( scanf( "%d%d%d", &A,&B,&C )!=EOF )
    {
        Graph G ;
        Table T ;
        found = false ;
        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( !found && 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( j==C )
            {
                //printf( "%d\n", curDist );
                PrintPath(T,i,j) ;
                printf("success\n") ;
                found = true ;
                break ;
            }

            //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( adjj==C )
                    {
                        //printf( "%d\n", curDist+1 );
                        PrintPath(T,adji,adjj) ;
                        printf("success\n") ;
                        found = true ;
                        break ;
                    }

                    //入队
                    ++size ;
                    ++rear ;
                    myQueue[rear].a = adji ;
                    myQueue[rear].b = adjj ;
                }
            }//for( k=0 ; k<6 ; ++k )
        }//while( !found && size != 0 )
    }//while( scanf( "%d%d%d", &A,&B,&C ) )
}