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 jis 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 operationsK. 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) 這道題的意思大家都能了解，就是給你三個空水杯，看你怎麼操作把第三個水杯裝滿。困擾了我很久的這道題，終于被我解決了，雖然程式長了一點，但是我還是蠻開心能解決它的。
利用string的增添字元友善，用字元去代表操作，然後最後周遊字元，将操作指令輸出。
#include<stdio.h>
#include<string>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int a,b,n;
string s;
};
int a,b,c;
int book[110][110];
void cn()
{
queue<node>Q;
node st,en;
st.a=a,st.b=0,st.n=1,st.s="1";
Q.push(st);
st.a=0,st.b=b,st.s="2";
Q.push(st);
book[a][0]=book[0][b]=1;
while(!Q.empty())
{
en=Q.front();
Q.pop();
if(en.a==c||en.b==c)
{
printf("%d\n",en.n);
for(int i=0; i<en.n; i++)
{
if(en.s[i]=='1')
printf("FILL(1)\n");
else if(en.s[i]=='2')
printf("FILL(2)\n");
else if(en.s[i]=='3')
printf("POUR(1,2)\n");
else if(en.s[i]=='4')
printf("POUR(2,1)\n");
else if(en.s[i]=='5')
printf("DROP(1)\n");
else if(en.s[i]=='6')
printf("DROP(2)\n");
}
return;
}
for(int i=1; i<=6; i++)
{
if(i==1)
{
if(en.a!=a&&book[a][en.b]==0)
{
st.a=a;
st.b=en.b;
st.s=en.s;
st.s+='1';
st.n=en.n+1;
book[a][en.b]=1;
Q.push(st);
}
}
else if(i==2)
{
if(en.b!=b&&book[en.a][b]==0)
{
st.a=en.a;
st.b=b;
st.s=en.s;
st.s+='2';
st.n=en.n+1;
book[en.a][b]=1;
Q.push(st);
}
}
else if(i==3)
{
if(en.a!=0&&en.b!=b)
{
if(en.a>(b-en.b))
{
st.a=en.a-(b-en.b);
st.b=b;
if(book[st.a][st.b]==0)
{
st.n=en.n+1;
st.s=en.s;
st.s+='3';
book[st.a][st.b]=1;
Q.push(st);
}
}
else
{
st.a=0;
st.b=en.b+en.a;
if(book[st.a][st.b]==0)
{
st.n=en.n+1;
st.s=en.s;
st.s+='3';
book[st.a][st.b]=1;
Q.push(st);
}
}
}
}
else if(i==4)
{
if(en.b!=0&&en.a!=a)
{
if(en.b>(a-en.a))
{
st.a=a;
st.b=en.b-(a-en.a);
if(book[st.a][st.b]==0)
{
st.n=en.n+1;
st.s=en.s;
st.s+='4';
book[st.a][st.b]=1;
Q.push(st);
}
}
else
{
st.a=en.a+en.b;
st.b=0;
if(book[st.a][st.b]==0)
{
st.n=en.n+1;
st.s=en.s;
st.s+='4';
book[st.a][st.b]=1;
Q.push(st);
}
}
}
}
else if(i==5)
{
if(en.a!=0&&book[0][en.b]==0)
{
st.a=0;
st.b=en.b;
st.n=en.n+1;
st.s=en.s;
st.s+='5';
book[0][en.b]=1;
Q.push(st);
}
}
else if(i==6)
{
if(en.b!=0&&book[en.a][0]==0)
{
st.a=en.a;
st.b=0;
st.n=en.n+1;
st.s=en.s;
st.s+='6';
book[en.a][0]=1;
Q.push(st);
}
}
}
}
printf("impossible\n");
}
int main()
{
while(~scanf("%d%d%d",&a,&b,&c)&&a+b+c)
{
memset(book,0,sizeof(book));
if(c>max(a,b))
printf("impossible\n");
else
cn();
}
return 0;
}
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