Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11705		Accepted: 4956		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

    题意：输入3个整数n,m,k，前两个整数代表两个杯子的容量。要求用两个杯子通过倒满。倒空。倒入还有一个杯子等方法使得两个杯子中的当中一个杯子的水量等于k，并输出步骤。

    思路：题目非常easy，就是麻烦。特别是步骤那一部分，须要用到BFS的回溯，通过BFS进行搜索，记录两个杯子的水量，使得后面不得反复。

假设搜不到就输出impossible

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            using namespace std;int n,m,k;int ans;int v[110][110];struct node{ int x; int y; int z; int cnt;} a[1000010];void DFS(int kk){ int pt = a[kk].cnt; if(pt<=0) { return ; } DFS(pt); if(a[pt].x == 1) { if(a[pt].y == 1) { printf("FILL(1)\n"); } else { printf("FILL(2)\n"); } } else if(a[pt].x == 2) { if(a[pt].y == 1) { printf("DROP(1)\n"); } else { printf("DROP(2)\n"); } } else if(a[pt].x == 3) { if(a[pt].y == 1) { printf("POUR(1,2)\n"); } else { printf("POUR(2,1)\n"); } }}void BFS(){ ans = 1; queue
            
             q; memset(v,0,sizeof(v)); struct node t,f; t.x = 0; t.y = 0; t.z = 0; t.cnt = 0; a[0].x = 0; a[0].y = 0; a[0].cnt = 0; q.push(t); v[t.x][t.y] = 1; while(!q.empty()) { t = q.front(); q.pop(); for(int i=1; i<=3; i++) { for(int j=1; j<=2; j++) { f.x = t.x; f.y = t.y; if(i == 1) { if(j == 1 && f.x!=n) { f.x = n; } else if(j == 2 && f.y!=m) { f.y = m; } } else if(i == 2) { if(j == 1 && f.x!=0) { f.x = 0; } else if(j == 2 && f.y!=0) { f.y = 0; } } else if(i == 3) { if(j == 1 && (f.x!=0 && f.y!=m)) { if(f.x>=m-f.y) { f.x = f.x - m + f.y; f.y = m; } else { f.y = f.y + f.x; f.x = 0; } } else if(j == 2 && (f.y!=0 && f.x!=n)) { if(f.y>=n-f.x) { f.y = f.y - n + f.x; f.x = n; } else { f.x = f.x + f.y; f.y = 0; } } } if(v[f.x][f.y] == 0) { f.cnt = ans; f.z = t.z + 1; a[ans].x = i; a[ans].y = j; a[ans].cnt = t.cnt; q.push(f); v[f.x][f.y] = 1; if(f.x == k || f.y == k) { printf("%d\n",f.z); DFS(ans); if(i == 1) { if(j == 1) { printf("FILL(1)\n"); } else { printf("FILL(2)\n"); } } else if(i == 2) { if(j == 1) { printf("DROP(1)\n"); } else { printf("DROP(2)\n"); } } else if(i == 3) { if(j == 1) { printf("POUR(1,2)\n"); } else { printf("POUR(2,1)\n"); } } return ; } ans++; } } } } printf("impossible\n");}int main(){ while(scanf("%d%d%d",&n,&m,&k)!=EOF) { BFS(); } return 0;}