poj3414 Pots

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)

這道題可以用寬搜做，思路挺簡單的，就是寫起來比較復雜，一個小錯誤讓我改了4個小時的代碼。。。

#include
#include
int a,b,k,num,e[111111];
int vis[105][105];
struct node
{
	int x,y,from,num,caozuo;   //from記錄這個點的父親，num表示執行了多少次，caozuo為1~6，每一個操作對應一種輸出 
}q[111111];
int bfs()
{
	int front=1,rear=1,x,xx,y,yy,i,flag;
	q[1].x=0;q[1].y=0;q[1].num=0;q[1].from=-1;q[1].caozuo=-1;vis[0][0]=1;
	while(front<=rear)
	{
		x=q[front].x;y=q[front].y;
		if(x==k || y==k){
			num=front;
			return q[front].num;
		}
		front++;
		if(x!=a){
			xx=a;yy=y;
			if(vis[xx][yy]==0)
			{vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=1;}
		}
		if(y!=b){
			xx=x;yy=b;
			if(vis[xx][yy]==0)
			{vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=2;}
		}
		if(x!=0){
			xx=0;yy=y;
            if(vis[xx][yy]==0)
			{vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=3;}
		}
		if(y!=0){
	        xx=x;yy=0;
            if(vis[xx][yy]==0)
			{vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=4;}
		}
		if(x!=0){
			if(yb){
			    xx=x+y-b;yy=b;
                if(vis[xx][yy]==0)
			   {vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=5;}
			 }
			}
		}
		if(y!=0){
			if(xa){
			 xx=a;yy=x+y-a;
             if(vis[xx][yy]==0)
			  {vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=6;}
			 }
			}
		}
	}
	return -1;
}


int main()
{
	int i,j,t,tt;
	while(scanf("%d%d%d",&a,&b,&k)!=EOF)
	{
		memset(vis,0,sizeof(vis));
		tt=bfs();
		if(tt==-1){
			printf("impossible\n");continue;
		}
		printf("%d\n",tt);
		t=0;
		while(num!=1)
		{
			e[++t]=q[num].caozuo;
			num=q[num].from;
		}
		for(i=t;i>=1;i--){
			if(e[i]==1)printf("FILL(1)\n");
			else if(e[i]==2)printf("FILL(2)\n");
			else if(e[i]==3)printf("DROP(1)\n");
			else if(e[i]==4)printf("DROP(2)\n");
			else if(e[i]==5)printf("POUR(1,2)\n");
			else if(e[i]==6)printf("POUR(2,1)\n");
		}
	}
	return 0;
}