POJ3414--Pots

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 potj is full (and there may be some water left in the pot i), or the poti 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 exactlyC liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, andC. 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 operationsK. 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)

/*C<=max(a,b)....a,b,c上限是100每一次可以进行六种操作：一：装满A 二：装满B三：倒掉A 四：倒掉B五：A倒B  六：B倒A判断是否达到目标的条件：在队列首判断va或者vb。如果等就可以退出队列了（原先我一直以为可以va+vb==c)贡献若干个WA至于要输出过程嘛。可以用一个结构体来存。*/
#include <iostream>#include <cstdio>#include <queue>#include <cstring>#include <string>using namespace std;#define maxn 108#define inf 0x3f3f3f3fint dis[maxn][maxn];string A[maxn*maxn];struct PRE{int prea,preb;bool fill;bool drop;bool pour;int v1,v2;//如果只需要用到一个的时候就直接用}pre[maxn][maxn];//pre[a][b]用来存第一个罐子的水量为a，第二个罐子的水量为b的前驱inline int man(int a,int b){return a>b?a:b;}int main(){int a,b,c;//a,b分别是容量，c=是要装出来的while(scanf("%d%d%d",&a,&b,&c)==3){bool flag=false;memset(dis,0x3f,sizeof(dis));int mab=max(a,b);for(int i=0;i<=mab;i++){for(int j=0;j<=mab;j++){dis[i][j]=inf;pre[i][j].drop=pre[i][j].fill=pre[i][j].pour=0;}}dis[0][0]=0;queue <int> qa;queue <int> qb;qa.push(0);qb.push(0);int ta,tb;while(!qa.empty()){int va=qa.front();int vb=qb.front();qa.pop();qb.pop();if(va==c||vb==c){ta=va;tb=vb;flag=true;break;}//操作一：fill aif(dis[va][vb]+1<dis[a][vb]){dis[a][vb]=dis[va][vb]+1;pre[a][vb].fill=true;pre[a][vb].drop=pre[a][vb].pour=false;pre[a][vb].v1=1;pre[a][vb].prea=va;pre[a][vb].preb=vb;qa.push(a);qb.push(vb);}//操作二：fill bif(dis[va][vb]+1<dis[va][b]){dis[va][b]=dis[va][vb]+1;pre[va][b].fill=true;pre[va][b].pour=pre[va][b].drop=false;pre[va][b].v1=2;pre[va][b].prea=va;pre[va][b].preb=vb;qa.push(va);qb.push(b);}//操作三：倒掉 aif(dis[va][vb]+1<dis[0][vb]){dis[0][vb]=dis[va][vb]+1;pre[0][vb].drop=true;pre[0][vb].fill=pre[0][vb].pour=false;pre[0][vb].v1=1;pre[0][vb].prea=va;pre[0][vb].preb=vb;qa.push(0);qb.push(vb);}//操作四：倒掉 bif(dis[va][vb]+1<dis[va][0]){dis[va][0]=dis[va][vb]+1;pre[va][0].drop=true;pre[va][0].fill=pre[va][0].pour=false;pre[va][0].v1=2;pre[va][0].prea=va;pre[va][0].preb=vb;qa.push(va);qb.push(0);}//操作五： a倒入bif(va+vb<=b&&dis[va][vb]+1<dis[0][va+vb]){dis[0][va+vb]=dis[va][vb]+1;pre[0][va+vb].pour=true;pre[0][va+vb].fill=pre[0][va+vb].drop=false;pre[0][va+vb].v1=1;pre[0][va+vb].v2=2;pre[0][va+vb].prea=va;pre[0][va+vb].preb=vb;qa.push(0);qb.push(va+vb);}if(va+vb>b&&dis[va][vb]+1<dis[va+vb-b][b]){dis[va+vb-b][b]=dis[va][vb]+1;pre[va+vb-b][b].pour=true;pre[va+vb-b][b].fill=pre[va+vb-b][b].drop=false;pre[va+vb-b][b].v1=1;pre[va+vb-b][b].v2=2;pre[va+vb-b][b].prea=va;pre[va+vb-b][b].preb=vb;qa.push(va+vb-b);qb.push(b);}//操作六： b倒入aif(va+vb<=a&&dis[va][vb]+1<dis[va+vb][0]){dis[va+vb][0]=dis[va][vb]+1;pre[va+vb][0].pour=true;pre[va+vb][0].fill=pre[va+vb][0].drop=false;pre[va+vb][0].v1=2;pre[va+vb][0].v2=1;pre[va+vb][0].prea=va;pre[va+vb][0].preb=vb;qa.push(va+vb);qb.push(0);}if(va+vb>a&&dis[va][vb]+1<dis[a][va+vb-a]){dis[a][va+vb-a]=dis[va][vb]+1;pre[a][va+vb-a].pour=true;pre[a][va+vb-a].fill=pre[a][va+vb-a].drop=false;pre[a][va+vb-a].v1=2;pre[a][va+vb-a].v2=1;pre[a][va+vb-a].prea=va;pre[a][va+vb-a].preb=vb;qa.push(a);qb.push(va+vb-a);}}if(!flag) cout<<"impossible"<<endl;else{cout<<dis[ta][tb]<<endl;//ta tbint sum=dis[ta][tb];int fuck=sum;int ia=ta,ib=tb;while(1){if(ia==0&&ib==0) break;if(pre[ia][ib].fill){A[fuck]="FILL(";A[fuck].push_back('0'+pre[ia][ib].v1);A[fuck].push_back(')');fuck--;}if(pre[ia][ib].drop){A[fuck]="DROP(";A[fuck].push_back('0'+pre[ia][ib].v1);A[fuck].push_back(')');fuck--;}if(pre[ia][ib].pour){A[fuck]="POUR(";A[fuck].push_back('0'+pre[ia][ib].v1);A[fuck].push_back(',');A[fuck].push_back('0'+pre[ia][ib].v2);A[fuck].push_back(')');fuck--;}int iia=ia,iib=ib;ia=pre[iia][iib].prea;ib=pre[iia][iib].preb;}for(int i=1;i<=sum;i++){cout<<A[i]<<endl;}}}return 0;}