一、題面

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)

二、分析

 對於這題，難點不在BFS的思路，難點在於BFS每一次父節點生成孩子結點的時候，情況比較複雜。對於記錄路徑，仍然需要使用孩子節點標記一個字首指向父節點，然後用遞迴的方式實現即可。自己在寫程式碼的時候非常不注意，在生成孩子節點時，對於標記訪問的陣列，本來應該用=，但我直接複製的判斷條件裡的==,導致一直RE。謹記！

三、AC程式碼

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 using namespace std;
  6 const int MAXN = 100;
  7 bool visit[MAXN+3][MAXN+3];
  8 int A, B, C;
  9 
 10 struct Point
 11 {
 12     int first, second, cnt;
 13     int prev, id;   //父節點和操作物件
 14     char op; //操作
 15 };
 16 
 17 Point P[MAXN*MAXN + 4];
 18 int Cnt, Ans;
 19 
 20 void Output(int t)
 21 {
 22     if(P[t].prev != -1)
 23     {
 24         Ans++;
 25         Output(P[t].prev);
 26     }
 27     if(Ans != -1)
 28     {
 29         printf("%d\n", Ans);
 30         Ans = -1;
 31     }
 32     if(P[t].op=='F')
 33     {
 34         printf("FILL(%d)\n", P[t].id);
 35     }
 36     else if(P[t].op == 'P')
 37     {
 38         printf("POUR(%d,%d)\n", P[t].id, P[t].id==1?2:1);
 39     }
 40     else if(P[t].op == 'D')
 41     {
 42         printf("DROP(%d)\n", P[t].id);
 43     }
 44 }
 45 
 46 void BFS()
 47 {
 48     Point t;
 49     int cur;
 50     t.first = 0, t.second = 0;
 51     visit[0][0] = 1;
 52     t.op = '0', t.prev = -1, t.id = -1;
 53     t.cnt = 0;
 54     P[0] = t;
 55     Cnt = 1;
 56     cur = 0;
 57 
 58     while(true)
 59     {
 60         if(cur >= Cnt)
 61         {
 62             printf("impossible\n");
 63             return;
 64         }
 65         Point pt = P[cur++];
 66 
 67         if(pt.first == C || pt.second == C)
 68         {
 69             Ans = 0;
 70             Output(pt.cnt);
 71             break;
 72         }
 73 
 74         t.prev = pt.cnt;
 75 
 76         if(pt.first < A)
 77         {
 78             t.first = A;
 79             t.second = pt.second;
 80             if(visit[t.first][t.second] == 0)
 81             {
 82                 //visit[t.first][t.second] == 1; 剛開始RE的原因
 83                 visit[t.first][t.second] = 1;
 84                 t.op = 'F';
 85                 t.id = 1;
 86                 t.cnt = Cnt;
 87                 P[Cnt++] = t;
 88 
 89             }
 90         }
 91 
 92         if(pt.second < B)
 93         {
 94             t.first = pt.first;
 95             t.second = B;
 96             if(visit[t.first][t.second] == 0)
 97             {
 98                 visit[t.first][t.second] = 1;
 99                 t.op = 'F';
100                 t.id = 2;
101                 t.cnt = Cnt;
102                 P[Cnt++] = t;
103             }
104         }
105 
106         if(pt.first < A && pt.second > 0 )
107         {
108             t.first = pt.first + pt.second;
109             t.second = t.first - A;
110             if(t.second < 0)
111                 t.second = 0;
112             else
113                 t.first = A;
114             if(visit[t.first][t.second] == 0)
115             {
116                 visit[t.first][t.second] = 1;
117                 t.op = 'P';
118                 t.id = 2;
119                 t.cnt = Cnt;
120                 P[Cnt++] = t;
121             }
122         }
123 
124         if(pt.second < B && pt.first > 0)
125         {
126             t.second = pt.second + pt.first;
127             t.first = t.second - B;
128             if(t.first < 0)
129                 t.first = 0;
130             else
131                 t.second = B;
132             if(visit[t.first][t.second] == 0)
133             {
134                 visit[t.first][t.second] = 1;
135                 t.op = 'P';
136                 t.id = 1;
137                 t.cnt = Cnt;
138                 P[Cnt++] = t;
139             }
140         }
141 
142         if(pt.first > 0)
143         {
144             t.first = 0;
145             t.second = pt.second;
146             if(visit[t.first][t.second] == 0)
147             {
148                 visit[t.first][t.second] = 1;
149                 t.op = 'D';
150                 t.id = 1;
151                 t.cnt = Cnt;
152                 P[Cnt++] = t;
153             }
154         }
155 
156         if(pt.second > 0)
157         {
158             t.first = pt.first;
159             t.second = 0;
160             if(visit[t.first][t.second] == 0)
161             {
162                 visit[t.first][t.second] = 1;
163                 t.op = 'D';
164                 t.id = 2;
165                 t.cnt = Cnt;
166                 P[Cnt++] = t;
167             }
168         }
169     }
170 }
171 
172 int main()
173 {
174     while(scanf("%d %d %d", &A, &B, &C)!=EOF)
175     {
176         memset(visit, 0, sizeof(visit));
177         BFS();
178     }
179     return 0;
180 }