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
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <queue> 5 #include <string> 6 using namespace std; 7 int A,B,C; 8 bool flag[101][101]; 9 char str[6][10]={"FILL(1)","FILL(2)","POUR(2,1)","POUR(1,2)","DROP(1)","DROP(2)"}; 10 struct node 11 { 12 int num1,num2; 13 string a; 14 int step; 15 }; 16 void bfs() 17 { 18 node p,temp; 19 int i; 20 p.num1=0; 21 p.num2=0; 22 p.step=0; 23 p.a=""; 24 queue<node>q; 25 while(!q.empty()) 26 q.pop(); 27 q.push(p); 28 while(!q.empty()) 29 { 30 p=q.front(); 31 q.pop(); 32 if(flag[p.num1][p.num2]) 33 continue; 34 else 35 { 36 flag[p.num1][p.num2]=true; 37 } 38 if(p.num1==C || p.num2==C) 39 { 40 printf("%d\n",p.step); 41 int j; 42 for(j=0;j<p.step;j++) 43 { 44 printf("%s\n",str[(p.a[j])-'0']); 45 } 46 return; 47 } 48 for(i=0;i<4;i++) 49 { 50 temp=p; 51 temp.step=p.step+1; 52 if(i==0) 53 { 54 if(temp.num1==0) 55 { 56 temp.num1=A; 57 temp.num2=p.num2; 58 temp.a=p.a+'0'; 59 q.push(temp); 60 } 61 if(temp.num2==0) 62 { 63 temp.num2=B; 64 temp.num1=p.num1; 65 temp.a=p.a+'1'; 66 q.push(temp); 67 } 68 } 69 if(i==1) 70 { 71 if(temp.num1!=0) 72 { 73 temp.num1=0; 74 temp.num2=p.num2; 75 temp.a=p.a+'4'; 76 q.push(temp); 77 } 78 if(temp.num2!=0) 79 { 80 temp.num2=0; 81 temp.num1=p.num1; 82 temp.a=p.a+'5'; 83 q.push(temp); 84 } 85 } 86 if(i==2) 87 { 88 if(temp.num1!=0 && temp.num2!=B) 89 { 90 int dd=B-temp.num2; 91 if(temp.num1>=dd) 92 { 93 temp.num2=B; 94 temp.num1-=dd; 95 } 96 else 97 { 98 temp.num2+=temp.num1; 99 temp.num1=0; 100 } 101 temp.a=p.a+'3'; 102 q.push(temp); 103 } 104 } 105 if(i==3) 106 { 107 if(temp.num2!=0 && temp.num1!=A) 108 { 109 int dd=A-temp.num1; 110 if(temp.num2>=dd) 111 { 112 temp.num1=A; 113 temp.num2-=dd; 114 } 115 else 116 { 117 temp.num1+=temp.num2; 118 temp.num2=0; 119 } 120 temp.a=p.a+'2'; 121 q.push(temp); 122 } 123 } 124 } 125 } 126 printf("impossible\n"); 127 } 128 int main() 129 { 130 while(~scanf("%d%d%d",&A,&B,&C)) 131 { 132 memset(flag,false,sizeof(flag)); 133 bfs(); 134 } 135 return 0; 136 }
BFS的思想；
题目大意：
给你两个空瓶子，只有三种操作
一、把一个瓶子灌满
二、把一个瓶子清空
三、把一个瓶子里面的水灌到另一个瓶子里面去（倒满之后要是还存在水那就依然在那个瓶子里面，或者被灌的瓶子有可能没满）
说实话，要是没有把这个题目放到bfs里面的话，真心不知道。之前直接是用模拟来实现，可是无法保证是最少操作数。
不过要是理解了bfs的思想来解这道题还是没什么压力的。