poj3414 倒水并输出过程

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)
思路：bfs爆搜。首先确定状态:两个桶中水的体积，采取行动的次数，当前状态是由6种操作中那种得来的。

关键：

1.实现6种操作。

2.要用vis记录数组，否则最后的结果不一定是采取行动次数最少。

3.如何回溯输出路径。（借助二维结构体）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define N 105

using namespace std;

typedef struct note {
int v1, v2;//桶1,2中的水的体积
int step;
int type;//1到6分别表示FILL(1),FILL(2),DROP(1),DROP(2),POUR(1,2),POUR(2,1)
}cup;

cup pre

;//二维结构体记录路径
cup cur, tmp;//用来记录队列，队首元素和下一个可能的元素
int a, b, c;//输入
int vis

;//两维分别表示，两个桶的状态

void choose(int i)
{
int sum;
switch (i)
{
case 1:
tmp.v1 = a;
break;
case 2:
tmp.v2 = b;
break;
case 3:
tmp.v1 = 0;
break;
case 4:
tmp.v2 = 0;
break;
case 5:
sum = tmp.v1 + tmp.v2;
if (sum <= b)//倒完
{
tmp.v1 = 0;
tmp.v2 = sum;
}
else
{
tmp.v1 -= b - tmp.v2;
tmp.v2 = b;
}
break;
case 6:
sum = tmp.v1 + tmp.v2;
if (sum <= a)//倒完
{
tmp.v1 = sum;
tmp.v2 = 0;
}
else
{
tmp.v2 -= a - tmp.v1;
tmp.v1 = a;
}
break;
}
}

void print(cup t)//回溯输出路径
{
if (pre[t.v1][t.v2].v1 != 0 || pre[t.v1][t.v2].v2 != 0)
print(pre[t.v1][t.v2]);
switch (t.type)
{
case 1: printf("FILL(1)\n"); break;
case 2: printf("FILL(2)\n"); break;
case 3: printf("DROP(1)\n"); break;
case 4: printf("DROP(2)\n"); break;
case 5: printf("POUR(1,2)\n"); break;
case 6: printf("POUR(2,1)\n"); break;
default : break;
}
}

int bfs()
{
//初始化队列
queue<cup>	q;
while (q.size())	q.pop();
//初始化初状态并入队
cup t;
t.v1 = 0; t.v2 = 0; t.step = 0;// t.type = -1;
q.push(t);
vis[0][0] = 1;
pre[0][0].v1 = 0; pre[0][0].v2 = 0;// pre[0][0].type = -1;

while (q.size())
{
cur = q.front(); q.pop();
if (cur.v1 == c || cur.v2 == c)//两桶中有一个水的体积达到c
{
printf("%d\n", cur.step);
return 1;
}
for (int i = 1; i <= 6; i++)//共6种情况
{
tmp = cur;
choose(i);//采取行动
if (!vis[tmp.v1][tmp.v2])
{
tmp.step++;
tmp.type = i;
q.push(tmp);
vis[tmp.v1][tmp.v2] = 1;
pre[tmp.v1][tmp.v2] = cur;//记录路径
}
}
}

return 0;
}

int main()
{
int ans;
while (~scanf("%d%d%d", &a, &b, &c))
{
memset(vis, 0, sizeof(vis));
ans = bfs();
if (ans)
print(cur);
else
printf("impossible\n");
}
return 0;
}