Poj 3414 Pots (BFS) (simple question)

Poj 3414 Pots (BFS) (simple question)

 

Pots

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:11551		Accepted:4900		Special Judge

 

Description

You are given two pots, having the volumeAAndBLiters respectively. The following operations can be saved med:

FILL (I) fill the pot I(1 ≤ I≤ 2) from the tap; DROP (I) empty the pot ITo the drain; POUR (I, j) pour from pot ITo pot J; After this operation either the pot JIs full (and there may be some water left in the pot I), Or the pot IIs 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 exactlyCLiters of water in one of the pots.

Input

On the first and only line are the numbersA,B, AndC. These are all integers in the range from 1 to 100 andC≤ Max (A,B).

Output

The first line of the output must contain the length of the sequence of operationsK. The followingKLines 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)

Question:

Two cups, one spoon, the amount of water in the two cups changes through the spoon or water pouring operation, find the minimum number of steps required to change the initial state to the final state.

Ideas:

Bfs traversal, each time there are six changes, from the initial state change to the minimum number of steps required for the final state.

Code:

 

#include 
 
  #include 
  
   #define MAX 10001typedef struct Node{int a,b,step,pre,flag;};int A,B,C;Node queue[MAX];bool visit[101][101];int path[MAX],index;void bfs(){memset(visit,false,sizeof(visit));int front=0,rear=0;Node cur,next;cur.a=0,cur.b=0,cur.pre=-1,cur.step=0;visit[0][0]=true;queue[rear++]=cur;while(front!=rear){cur=queue[front++];if(cur.a==C||cur.b==C) break;for(int i=0;i<6;i++){switch(i){case 0: next.a=A; next.b=cur.b; next.flag=0; break;case 1:next.a=cur.a; next.b=B; next.flag=1; break;case 2:next.a=0; next.b=cur.b; next.flag=2; break;case 3:next.a=cur.a; next.b=0; next.flag=3; break;case 4:if(B-cur.b
   
    =0;){path[index++]=i;i=queue[i].pre;}printf(%d,queue[front-1].step);for(int i=index-1;i>=0;i--){switch(queue[path[i]].flag){case 0:printf(FILL(1));break;case 1:printf(FILL(2)); break;case 2:printf(DROP(1)); break;case 3:printf(DROP(2)); break;case 4:printf(POUR(1,2)); break;case 5:printf(POUR(2,1)); break;}}}int main(){scanf(%d %d %d,&A,&B,&C);bfs();return 0;}