Pots bfs

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 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)


bfs,六种操作，代码中op数组给出，然后需要标记ab，同样的状态不重复记录。queue不太会用，还是用数组队列吧


代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <cmath>
#include <string>
using namespace std;
int a,b,c,ta,tb,flag,head=0,tail=0;
string op[6]={"FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
struct que
{
    int al,bl,f,c;
    int ope;
}q[10000];
int vis[101][101];///用这个有保障
void print(int x)
{
    if(q[x].f!=-1)print(q[x].f);
    else return;
    cout<<op[q[x].ope]<<endl;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    q[tail].al=0,q[tail].bl=0,q[tail].f=-1,q[tail++].c=0;
    vis[0][0]=0;
    if(c)
    while(head<tail)
    {
        if(q[head].al==c||q[head].bl==c)
        {
            flag=1;
            cout<<q[head].c<<endl;
            print(head);
            break;
        }
        for(int i=0;i<6;i++)
        {
            flag=0;
            ta=q[head].al;
            tb=q[head].bl;
            if(i==0){ta=a;}
            else if(i==1){tb=b;}
            else if(i==2){ta=0;}
            else if(i==3){tb=0;}
            else if(i==4){(b-tb>=ta)?(tb+=ta,ta=0):(ta-=b-tb,tb=b);}
            else{a-ta>=tb?(ta+=tb,tb=0):(tb-=a-ta,ta=a);}
            if(vis[ta][tb])continue;
            vis[ta][tb]=1;
            q[tail].ope=i;
            q[tail].al=ta,q[tail].bl=tb,q[tail].f=head,q[tail++].c=q[head].c+1;
        }
        head++;
    }
    else {flag=1;cout<<0<<endl;}
    if(flag==0)cout<<"impossible"<<endl;
}