POJ 3414 Pots
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 13547 | Accepted: 5718 | 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
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
1 /*和之前的那道倒水题目十分相似，唯一的难点就是输出倒水的方式，我们可以记录每一个搜到的状态的前一状态在队列中的编号和，该状态与前一状态的关系，再递归输出结果即可*/ 2 #include<iostream> 3 using namespace std; 4 #include<cstdio> 5 #include<cstring> 6 #define N 10010 7 int head=-1,tail=-1; 8 int a,b,c; 9 bool flag[101][101]={0}; 10 struct node{ 11 int a1,b1,pre,dis,relat; 12 }que[N]; 13 int bfs() 14 { 15 while(head<tail) 16 { 17 ++head; 18 if(que[head].a1==c||que[head].b1==c) 19 { 20 return head; 21 } 22 if(!flag[que[head].a1][0]) 23 { 24 flag[que[head].a1][0]=true; 25 node nex; 26 nex.a1=que[head].a1;nex.b1=0; 27 nex.dis=que[head].dis+1; 28 nex.pre=head; 29 nex.relat=4; 30 ++tail; 31 que[tail]=nex; 32 } 33 if(!flag[0][que[head].b1]) 34 { 35 flag[0][que[head].b1]=true; 36 node nex; 37 nex.a1=0;nex.b1=que[head].b1; 38 nex.dis=que[head].dis+1; 39 nex.pre=head; 40 nex.relat=3; 41 ++tail; 42 que[tail]=nex; 43 } 44 if(!flag[que[head].a1][b]) 45 { 46 flag[que[head].a1][b]=true; 47 node nex; 48 nex.a1=que[head].a1;nex.b1=b; 49 nex.dis=que[head].dis+1; 50 nex.pre=head; 51 nex.relat=2; 52 ++tail; 53 que[tail]=nex; 54 } 55 if(!flag[a][que[head].b1]) 56 { 57 flag[a][que[head].b1]=true; 58 node nex; 59 nex.a1=a;nex.b1=que[head].b1; 60 nex.dis=que[head].dis+1; 61 nex.pre=head; 62 nex.relat=1; 63 ++tail; 64 que[tail]=nex; 65 } 66 if(que[head].a1>=(b-que[head].b1)&&!flag[que[head].a1-(b-que[head].b1)][b]) 67 { 68 flag[que[head].a1-(b-que[head].b1)][b]=true; 69 node nex; 70 nex.a1=que[head].a1-(b-que[head].b1);nex.b1=b; 71 nex.dis=que[head].dis+1; 72 nex.pre=head; 73 nex.relat=5; 74 ++tail; 75 que[tail]=nex; 76 } 77 if(que[head].a1<(b-que[head].b1)&&!flag[0][que[head].a1+que[head].b1]) 78 { 79 flag[0][que[head].a1+que[head].b1]=true; 80 node nex; 81 nex.a1=0;nex.b1=que[head].a1+que[head].b1; 82 nex.dis=que[head].dis+1; 83 nex.pre=head; 84 nex.relat=5; 85 ++tail; 86 que[tail]=nex; 87 } 88 if(que[head].b1>=(a-que[head].a1)&&!flag[a][que[head].b1-(a-que[head].a1)]) 89 { 90 flag[a][que[head].b1-(a-que[head].a1)]=true; 91 node nex; 92 nex.a1=a;nex.b1=que[head].b1-(a-que[head].a1); 93 nex.dis=que[head].dis+1; 94 nex.pre=head; 95 nex.relat=6; 96 ++tail; 97 que[tail]=nex; 98 } 99 if(que[head].b1<(a-que[head].a1)&&!flag[que[head].a1+que[head].b1][0]) 100 { 101 flag[que[head].a1+que[head].b1][0]=true; 102 node nex; 103 nex.a1=que[head].a1+que[head].b1;nex.b1=0; 104 nex.dis=que[head].dis+1; 105 nex.pre=head; 106 nex.relat=6; 107 ++tail; 108 que[tail]=nex; 109 } 110 } 111 return -1; 112 } 113 void out(int temp) 114 { 115 if(que[que[temp].pre].pre!=-1) 116 out(que[temp].pre); 117 if(que[temp].relat==1) 118 printf("FILL(1) "); 119 else if(que[temp].relat==2) 120 printf("FILL(2) "); 121 else if(que[temp].relat==3) 122 printf("DROP(1) "); 123 else if(que[temp].relat==4) 124 printf("DROP(2) "); 125 else if(que[temp].relat==5) 126 printf("POUR(1,2) "); 127 else printf("POUR(2,1) "); 128 } 129 /* 130 FILL(2) 131 POUR(2,1) 132 DROP(1) 133 POUR(2,1) 134 FILL(2) 135 POUR(2,1) 136 */ 137 int main() 138 { 139 scanf("%d%d%d",&a,&b,&c); 140 ++tail; 141 que[tail].a1=0;que[tail].b1=0; 142 que[tail].dis=0;que[tail].pre=-1; 143 flag[0][0]=true; 144 int temp=bfs(); 145 if(temp==-1) 146 printf("impossible "); 147 else { 148 printf("%d ",que[head].dis); 149 out(temp); 150 } 151 return 0; 152 }