Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12198		Accepted: 5147		Special Judge

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 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)

Source

, Western Subregion

 

空杯子倒水问题，这题比较简单，只有两个杯子，最初的时候都是空的，要倒出指定量的水有三种操作：

1、FILL(i)        把第i个杯子装满（i=0,1）

2、DROP(i)      把第i个杯子倒空

3、POUR(i,j)    把i的水倒入到j中，直到j满或i倒完

我的想法：把a->b，b->c，。。。。共6种倒水方法一个一个列出来，而每种都是一样的讨论方法，

虽然很好做，但对于我这样的入门级水手来说还是写不出太好看的代码，所以。。。。

看看吧，看不懂再去看看网上别人的==||

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;const int maxn=500;bool vis[maxn][maxn];int fa[maxn][maxn];int op[maxn][maxn];int sa,sb,sc;struct node{   int a,b,step;};char ans[6][20]={
   "FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};void print(int a ,int b){     if(op[a][b]==-1)     return;     else{         print(fa[a][b]/1000,fa[a][b]%1000);         printf("%s\n",ans[op[a][b]]);     }}bool bfs(){   node u,v;   u.a=0,u.b=0;   u.step=0;   queue
          
           q; q.push(u); memset(op,0,sizeof(op)); memset(vis,false,sizeof(vis)); memset(fa,0,sizeof(fa)); op[0][0]=-1; fa[0][0]=0; vis[0][0]=true; while(!q.empty()){ u=q.front(); q.pop(); if(u.a==sc||u.b==sc){ printf("%d\n",u.step); print(u.a,u.b); return true; } v=u; if(u.a!=sa){ v.a=sa; if(!vis[v.a][v.b]){ v.step++; q.push(v); op[v.a][v.b]=0; vis[v.a][v.b]=true; fa[v.a][v.b]=u.a*1000+u.b; } }v=u; if(u.b!=sb){v.b=sb; if(!vis[v.a][v.b]){ v.step=u.step+1; q.push(v); op[v.a][v.b]=1; vis[v.a][v.b]=1; fa[v.a][v.b]=u.a*1000+u.b; } }v=u; if(u.a){v.a=0; if(!vis[v.a][v.b]){ v.step++; q.push(v); op[v.a][v.b]=2; vis[v.a][v.b]=true; fa[v.a][v.b]=u.a*1000+u.b; } }v=u; if(u.b){v.b=0; if(!vis[v.a][v.b]){ v.step++; q.push(v); op[v.a][v.b]=3; vis[v.a][v.b]=true; fa[v.a][v.b]=u.a*1000+u.b; } }v=u; if(u.a){ if(v.a>=sb-u.b&&u.b!=sb){v.a-=(sb-u.b);v.b=sb;} else if(v.a
           
            =sa-u.a&&u.a!=sa){v.b-=(sa-u.a);v.a=sa;} else if(v.b