| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 8266 | Accepted: 3507 | 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) 题意：给定三个数A B C，前两个数代表容量为A和B的容器，有三种操作，FILL,DROP，POUR；求最少经过几次这样的操作可以得到容量为C的水； 思路：因为每个容器都有FILL,DROP，POUR三种操作，将两个容器的状态用队列来维护，两个容器共有六种操作的可能；
另一个关键是打印路径，可以用一个结构体数组保存前驱，通过栈再将路径打印出来；
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
//各种状态
#define FILL_A 1;
#define FILL_B 2;
#define DROP_A 3;
#define DROP_B 4;
#define POURA_B 5;
#define POURB_A 6;
using namespace std; struct node
{
int x,y;
int step;
};
queue<struct node>que; struct Path
{
int x,y;
int way;
}path[][];//保存前驱； int vis[][];
int a,b,c; //打印路径
void Print_Path(int x, int y)
{
stack<struct Path>st;
while(!st.empty())
st.pop(); while(!(x == && y == ))
{
st.push(path[x][y]);
int sx = path[x][y].x;
int sy = path[x][y].y;
x = sx;
y = sy;
} while(!st.empty())
{
switch(st.top().way)
{
case :printf("FILL(1)\n");break;
case :printf("FILL(2)\n");break;
case :printf("DROP(1)\n");break;
case :printf("DROP(2)\n");break;
case :printf("POUR(1,2)\n");break;
case :printf("POUR(2,1)\n");break;
}
st.pop();
}
} void bfs()
{
while(!que.empty())
que.pop();
que.push((struct node){,,});//初始状态进队列
vis[][] = ;
while(!que.empty())
{
struct node u = que.front();
que.pop();
if(u.x == c || u.y == c)
{
printf("%d\n",u.step);
Print_Path(u.x,u.y);
return;
}
//FILL_A
if(u.x < a && !vis[a][u.y])
{
vis[a][u.y] = ;
que.push((struct node){a,u.y,u.step+});
path[a][u.y] = (struct Path){u.x,u.y,};
}
//FILL_B
if(u.y < b && !vis[u.x][b])
{
vis[u.x][b] = ;
que.push((struct node){u.x,b,u.step+});
path[u.x][b] = (struct Path){u.x,u.y,};
}
//DROP_A
if(u.x > && !vis[][u.y])
{
vis[][u.y] = ;
que.push((struct node){,u.y,u.step+});
path[][u.y] = (struct Path){u.x,u.y,};
}
//DROP_B
if(u.y > && !vis[u.x][])
{
vis[u.x][] = ;
que.push((struct node){u.x,,u.step+});
path[u.x][] = (struct Path){u.x,u.y,};
}
//POURA_B
if(u.x < a && u.y > )
{
int tmp = min(a-u.x,u.y);
if(!vis[u.x+tmp][u.y-tmp])
{
vis[u.x+tmp][u.y-tmp] = ;
que.push((struct node){u.x+tmp,u.y-tmp,u.step+});
path[u.x+tmp][u.y-tmp] = (struct Path){u.x,u.y,};
}
}
//POURB_A
if(u.x > && u.y < b)
{
int tmp = min(u.x,b-u.y);
if(!vis[u.x-tmp][u.y+tmp])
{
vis[u.x-tmp][u.y+tmp] = ;
que.push((struct node){u.x-tmp,u.y+tmp,u.step+});
path[u.x-tmp][u.y+tmp] = (struct Path){u.x,u.y,};
}
}
}
printf("impossible\n");
}
int main()
{
scanf("%d %d %d",&a,&b,&c);
memset(vis,,sizeof(vis));
bfs();
return ;
}