poj 3414 Pots （bfs+线索）-白红宇

poj 3414 Pots （bfs+线索）

阅读量：6431 次

发布时间：2019-06-23

本文共 3570 字，大约阅读时间需要 11 分钟。

Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10071		Accepted: 4237		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

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

思路：一共同拥有6种操作：把A中水倒掉，把A加满，把B里的水倒入A中。B和A类似。

罐子最大容积为100，设一个常量N=100，开一个二维数组记录状态变化的值。

1、从水龙头往A里加水t,记录-t,

2、从水龙头往B里加水t,记录-t-N,

3、从B里面往A加水t，记录t

4、从A里面往B加水t。记录N+t

5、把A里水倒掉，记录2*N+t,(A原有水t)

6、把B里水倒掉，记录3*N+t,(B原有水t)

#include
     
      #include
      
       #include
       #include
        
         #include
         
          using namespace std;#define N 105const int inf=0x1f1f1f1f;int a,b,c,flag;int mark[N][N];struct node{    int x,y,t;    friend bool operator<(node a,node b)    {        return a.t>b.t;    }};void prif(int x,int y)       //递归输出路径{    if(x==0&&y==0)        return ;    if(mark[x][y]>3*N)    {        prif(x,mark[x][y]-3*N);        printf("DROP(2)\n");    }    else if(mark[x][y]>2*N)    {        prif(mark[x][y]-2*N,y);        printf("DROP(1)\n");    }    else if(mark[x][y]>N)    {        int tmp=mark[x][y]-N;        prif(x+tmp,y-tmp);        printf("POUR(1,2)\n");    }    else if(mark[x][y]>0)    {        int tmp=mark[x][y];        prif(x-tmp,y+tmp);        printf("POUR(2,1)\n");    }    else if(mark[x][y]>-N)    {        int tmp=-mark[x][y];        prif(x-tmp,y);        printf("FILL(1)\n");    }    else if(mark[x][y]<-N)    {        int tmp=N+mark[x][y];        prif(x,y+tmp);        printf("FILL(2)\n");    }}void bfs(){    priority_queue
          
           q; node cur,next; mark[0][0]=inf; //该状态仅仅能出现一次。赋值为inf防止干扰其它值 mark[a][0]=-a; mark[0][b]=-b-N; cur.t=1; cur.x=a; cur.y=0; q.push(cur); cur.x=0; cur.y=b; q.push(cur); while(!q.empty()) { cur=q.top(); q.pop(); next.t=cur.t+1; if(cur.x==c||cur.y==c) { flag=1; printf("%d\n",cur.t); prif(cur.x,cur.y); return ; } if(cur.x
           
            0) //来自B的水 { int tmp=min(cur.y,a-cur.x); next.x=cur.x+tmp; next.y=cur.y-tmp; if(!mark[next.x][next.y]) { mark[next.x][next.y]=tmp; q.push(next); } } } if(cur.y
            
             0) //来自A的水 { int tmp=min(cur.x,b-cur.y); next.y=cur.y+tmp; next.x=cur.x-tmp; if(!mark[next.x][next.y]) { mark[next.x][next.y]=tmp+N; q.push(next); } } } if(cur.x>0) //倒掉水 { int tmp=cur.x; next.x=0; next.y=cur.y; if(!mark[next.x][next.y]) { mark[next.x][next.y]=2*N+tmp; q.push(next); } } if(cur.y>0) { int tmp=cur.y; next.y=0; next.x=cur.x; if(!mark[next.x][next.y]) { mark[next.x][next.y]=3*N+tmp; q.push(next); } } }}int main(){ while(scanf("%d%d%d",&a,&b,&c)!=-1) { memset(mark,0,sizeof(mark)); flag=0; bfs(); if(!flag) printf("impossible\n"); } return 0;}

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