描述
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.
輸入
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).
輸出
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’.
樣例輸入
3 5 4
樣例輸出
6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)
題目大意：有2個pots，A，B分别表示2個pots的最大容積，要求通過幾種操作方式，得到輸入的C所需的容積。
代碼部分:
#include <iostream>
#include <cstring>
#include <stdio.h>
using namespace std;
int visited[101][101];
int A,B,C;
struct Node{
int a,b,opeat,parent,step;
Node(int x,int y,int o,int p,int s):a(x),b(y),opeat(o),parent(p),step(s){}
Node(){}
}que[10001];
int head=0,tail=0,step=0;
int opeator;
void print(int x){ //輸出
if(que[x].parent>=0)
print(que[x].parent);
switch(que[x].opeat){
case 0:{
printf("FILL(1)\n");
break;
}case 1:{
printf("FILL(2)\n");
break;
}case 2:{
printf("DROP(1)\n");
break;
}case 3:{
printf("DROP(2)\n");
break;
}case 4:{
printf("POUR(1,2)\n");
break;
}case 5:{
printf("POUR(2,1)\n");
break;
}default:
break;
}
}
void Bfs(int a,int b,int c){
if(a==0&&b==0&&step==0){
//visited[A][B]=1;
que[head]=Node(0,0,-1,-1,step);
visited[a][b]=1;
step++;
}
if(a==c||b==c){
//cout<<"head:";
cout<<que[head].step<<endl;
print(head);
return;
}
for(int i=0;i<6;i++){ //周遊6種方法
tail++;
switch(i){
case 0:{//FILL(1)
if(a<A&&!visited[A][b]){
visited[A][b]=1;
opeator=0;
que[tail]=Node(A,b,opeator,head,que[head].step+1);
}else
tail--;
break;
}case 1:{//FILL(2)
if(b<B&&!visited[a][B]){
visited[a][B]=1;
opeator=1;
que[tail]=Node(a,B,opeator,head,que[head].step+1);
}else
tail--;
break;
}case 2:{//DROP(1)
if(a!=0&&!visited[0][b]){
visited[0][b]=1;
opeator=2;
que[tail]=Node(0,b,opeator,head,que[head].step+1);
}else
tail--;
break;
}case 3:{//DROP(2)
if(b!=0&&!visited[a][0]){
visited[a][0]=1;
opeator=3;
que[tail]=Node(a,0,opeator,head,que[head].step+1);
}else
tail--;
break;
}case 4:{//POUR(1,2)
opeator=4;
if(a>=(B-b)&&!visited[a-B+b][B]&&a!=0){
visited[a-B+b][B]=1;
que[tail]=Node(a-B+b,B,opeator,head,que[head].step+1);
}else if(a<(B-b)&&!visited[0][b+a]&&a!=0){
visited[0][b+a]=1;
que[tail]=Node(0,b+a,opeator,head,que[head].step+1);
}else
tail--;
break;
}case 5:{//POUR(2,1)
opeator=5;
if(b>=(A-a)&&!visited[A][b-A+a]&&b!=0){
visited[A][b-A+a]=1;
que[tail]=Node(A,b-A+a,opeator,head,que[head].step+1);
}else if(b<(A-a)&&!visited[b+a][0]&&b!=0){
visited[b+a][0]=1;
que[tail]=Node(b+a,0,opeator,head,que[head].step+1);
}else
tail--;
break;
}default:
break;
}
if(que[tail].a==c||que[tail].b==c){
cout<<que[tail].step<<endl;
print(tail);
return;
}
}
head++;
if(head!=tail){
Bfs(que[head].a,que[head].b,c);
head++;
}
}
int main()
{
cin>>A>>B>>C;
memset(visited,0,sizeof(visited));
Bfs(0,0,C);
return 0;
}
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