poj 3414 (POTS) (BFS)-白红宇

Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12295		Accepted: 5190		Special Judge

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

6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)POUR(2,1)

Source

, Western Subregion

题目大意：

就是给你两个瓶子a b，看是否能够通过以下的步骤达到k

1)FILL(i):将 i 瓶子倒满水;

2)POUR(i,j):将i瓶子中的水倒入j

3)DROP(I):将i瓶子中的水倒出来：

/**解题思路：就是一个简单的BFS,仔细搞搞就行了**//**2015 - 09 - 16 下午Author: ITAKMotto:今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。**/#include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 using namespace std;typedef long long LL;const int maxn = 105;const double eps = 1e-7;bool vis[maxn][maxn];const int dir[4][2]= {1, 0, 0, 1, -1, 0, 0, -1};char map[maxn][maxn];int a, b, k;struct node{ int vola, volb, step; char str[maxn][maxn];};bool bfs(){ memset(vis, false, sizeof(vis)); queue
                 
                   que; node p, q; p.vola = 0, p.volb = 0, p.step = 0; que.push(p); vis[0][0] = 1; ///vis[p.vola][p.volb] = true; while(!que.empty()) { p = que.front(); que.pop(); if(p.vola==k || p.volb == k) { cout<
                  
                   <
                   
                     b if(p.volb < b) { q = p; if(q.volb + q.vola <= b) { q.volb += q.vola; q.vola = 0; } else { q.vola = (q.vola+q.volb)-b; q.volb = b; } q.step++; strcpy(q.str[q.step], "POUR(1,2)"); if(!vis[q.vola][q.volb]) { vis[q.vola][q.volb] = true; que.push(q); } } } ///把 b 倒满 if(p.volb == 0) { q = p; q.volb = b; q.step++; strcpy(q.str[q.step], "FILL(2)"); if(!vis[q.vola][q.volb]) { vis[q.vola][q.volb] = true; que.push(q); } } ///把 b 中的水倒出来 else if(p.volb <= b) { q = p; q.volb = 0; q.step++; strcpy(q.str[q.step],"DROP(2)"); if(!vis[q.vola][q.volb]) { vis[q.vola][q.volb] = true; que.push(q); } if(p.vola < a) { q = p; if(q.vola + q.volb <= a) { q.vola += q.volb; q.volb = 0; } else { q.volb = (q.vola+q.volb)-a; q.vola = a; } q.step++; strcpy(q.str[q.step], "POUR(2,1)"); if(!vis[q.vola][q.volb]) { vis[q.vola][q.volb] = true; que.push(q); } } } } return false;}int main(){ while(cin>>a>>b>>k) { bool ok = bfs(); if(!ok) puts("impossible"); } return 0;}/**Sample Input3 5 4Sample Output6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)POUR(2,1)**/