Pots(简单BFS+找前缀路径)
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题目描述:
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
- FILL(i) fill the poti(1 ≤i ≤ 2) from the tap;
- DROP(i) empty the potito the drain;
- POUR(i,j) pour from potito potj; after this operation either the potjis full (and there may be some water left in the poti), or the potiis empty (and all its contents have been moved to the potj).
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)
代码如下:
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
#define mm(a) memset(a,0,sizeof(a))
int A,B,C;
void print_path(int n)
{
switch(n)
{
case 0:
cout<<"FILL(1)";
break;
case 1:
cout<<"FILL(2)";
break;
case 2:
cout<<"DROP(1)";
break;
case 3:
cout<<"DROP(2)";
break;
case 4:
cout<<"POUR(1,2)";
break;
case 5:
cout<<"POUR(2,1)";
break;
}
cout<<endl;
}
struct node
{
int x,y,step,op;
} vis[110][110];
struct point
{
int x, y;
point(int xx,int yy)
{
x=xx,y=yy;
}
};
void DFS(int x, int y)//dfs找前缀
{
if(x==0&&y==0)
return;
DFS(vis[x][y].x, vis[x][y].y);
print_path(vis[x][y].op);
}
int BFS(point t)
{
queue<point>Q;
vis[0][0].step=1;
Q.push(t);
while(Q.size())
{
point s=Q.front();
Q.pop();
if(s.x==C||s.y==C)
{
cout<<vis[s.x][s.y].step-1<<endl;;
DFS(s.x,s.y);
return 0;
}
for(int i=0;i<6;i++)
{
point q=s;
if(i==0)//FILL(1)
q.x = A;
else if(i==1)//FILL(2)
q.y=B;
else if(i==2)//DROP(1)
q.x =0;
else if(i==3)//DROP(2)
q.y=0;
else if(i==4)//POUR(1,2)
{
if(q.x+q.y<=B)
q.y+=q.x, q.x=0;
else
q.x-=(B-q.y),q.y=B;
}
else//POUR(2,1)
{
if(q.x+q.y<=A)
q.x+=q.y,q.y=0;
else
q.y-=(A-q.x),q.x=A;
}
if( vis[q.x][q.y].step==0)
{
vis[q.x][q.y].step=vis[s.x][s.y].step+1;
vis[q.x][q.y].x=s.x;
vis[q.x][q.y].y=s.y;
vis[q.x][q.y].op=i;
Q.push(q);
}
}
}
return -1;
}
int main()
{
while(cin>>A>>B>>C)
{
mm(vis);
int ans=BFS(point(0,0));
if(ans==-1)
cout<<"impossible\n";
}
return 0;
}