Problem 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)
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bfs（），简单的，记录路径很重要
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1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cstdio> 5 #include<vector> 6 using namespace std; 7 int a,b,c; 8 int vis[1001][1001],step1,lastv;//vis标记数组 9 int id[10001]; 10 int tail,head,flag; 11 struct node 12 { 13 int pre;//记录前驱 14 int k1,k2;//两个桶中的水量 15 int operation;//操作 16 int step;//记录步数 17 }que[10000001],now,next; 18 void bfs() 19 { 20 tail=head=0; 21 now.k1=0; 22 now.k2=0; 23 now.pre=0; 24 now.operation=0; 25 now.step=0; 26 que[tail++]=now; 27 vis[0][0]=1; 28 while(head<tail) 29 { 30 now=que[head]; 31 head++; 32 if(now.k1==c||now.k2==c) 33 { 34 flag=1; 35 step1=now.step; 36 lastv=head-1; 37 return; 38 } 39 for(int it=0;it<6;it++)//六种操作 40 { 41 if(it==0)//fill(1); 42 { 43 next.k1=a; 44 next.k2=now.k2; 45 } 46 if(it==1)//fill(2); 47 { 48 next.k1=now.k1; 49 next.k2=b; 50 } 51 if(it==2)//drop(1); 52 { 53 next.k1=0; 54 next.k2=now.k2; 55 } 56 if(it==3)//drop(2); 57 { 58 next.k1=now.k1; 59 next.k2=0; 60 } 61 if(it==4)//pour(1,2); 62 { 63 if(now.k1+now.k2<=b) 64 { 65 next.k1=0; 66 next.k2=now.k1+now.k2; 67 } 68 else 69 { 70 next.k1=now.k1+now.k2-b; 71 next.k2=b; 72 } 73 } 74 if(it==5)//pour(2,1); 75 { 76 if(now.k1+now.k2<=a) 77 { 78 next.k1=now.k1+now.k2; 79 next.k2=0; 80 } 81 else 82 { 83 next.k2=now.k1+now.k2-a; 84 next.k1=a; 85 } 86 } 87 next.operation=it; 88 if(!vis[next.k1][next.k2])//标记检测完退出 89 { 90 vis[next.k1][next.k2]=1; 91 next.step=now.step+1; 92 next.pre=head-1; 93 //cout<<next.pre<<endl; 94 que[tail].k1=next.k1;que[tail].k2=next.k2; 95 que[tail].step=next.step;que[tail].pre=next.pre; 96 que[tail].operation=next.operation; 97 tail++; 98 } 99 if(next.k1==c||next.k2==c) 100 { 101 flag=1; 102 step1=next.step; 103 lastv=tail-1; 104 return; 105 106 } 107 } 108 109 } 110 } 111 int main() 112 { 113 while(scanf("%d%d%d",&a,&b,&c)!=EOF) 114 { 115 flag=0; 116 step1=0; 117 memset(vis,0,sizeof(vis)); 118 bfs(); 119 if(flag) 120 { 121 printf("%d\n",step1); 122 id[step1]=lastv; 123 for(int i=step1-1;i>=1;i--) 124 { 125 //printf("que[id[%d+1]].pre: %d\n",que[id[i+1]].pre); 126 id[i]=que[id[i+1]].pre; 127 } 128 for(int i=1;i<=step1;i++) 129 { 130 //printf("id[%d]:%d\n",i,id[i]); 131 if(que[id[i]].operation==0) 132 printf("FILL(1)\n"); 133 if(que[id[i]].operation==1) 134 printf("FILL(2)\n"); 135 if(que[id[i]].operation==2) 136 printf("DROP(1)\n"); 137 if(que[id[i]].operation==3) 138 printf("DROP(2)\n"); 139 if(que[id[i]].operation==4) 140 printf("POUR(1,2)\n"); 141 if(que[id[i]].operation==5) 142 printf("POUR(2,1)\n"); 143 } 144 } 145 else 146 printf("impossible\n"); 147 } 148 return 0; 149 }
转载于:https://www.cnblogs.com/sdau--codeants/p/3365030.html