题目：
| Pots| Time Limit: 1000MS | | Memory Limit: 65536K | | |
| ---------------------------- | - | ------------------------ | - | ------------- |
| Total Submissions: 15275 | | Accepted: 6433 | | Special Judge |DescriptionYou are given two pots, having the volume of A and B liters respectively. The following operations can be performed:1. FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
2. DROP(i) empty the pot i to the drain;
3. 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.InputOn 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).OutputThe 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 Input3 5 4Sample Output```
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
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思路：
给你两个容器，分别能装下A升水和B升水，并且可以进行以下操作\
FILL(i) 将第i个容器从水龙头里装满(1 ≤ i ≤ 2);\
DROP(i) 将第i个容器抽干\
POUR(i,j) 将第i个容器里的水倒入第j个容器（这次操作结束后产生两种结果，一是第j个容器倒满并且第i个容器依旧有剩余，二是第i个容器里的水全部倒入j中，第i个容器为空）\
现在要求你写一个程序，来找出能使其中任何一个容器里的水恰好有C升，找出最少操作数并给出操作过程
\
运用啦数组模拟队列的方法，这样方便记录路径，总共六种情况：
1.装满1
2.装满2
3.倒掉1
4.倒掉2
5.把1倒到2（分类讨论）
6.把2倒到1（分类讨论）
代码;
```cpp
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int vis[110][110];
int a,b,c;
int step;
int flag;
//记录状态
struct node
{
int k1,k2;//当前水杯里面的水的量
int op;//当前的操作
int step;//步数
int pre;//前一步的下标
} q[20000];
int id[20000];//最终操作在队列的编号
int lastindex;//最后一个的标号
void bfs()
{
node now,to;
int head,tail;
head=tail=0;
q[tail].k1=0;
q[tail].k2=0;
q[tail].op=0;
q[tail].step=0;
q[tail].pre=0;
tail++;
mem(vis,0);
//以上操作为初始化
vis[0][0]=1;//初始状态入队
while(head<tail)//队列非空时
{
now=q[head];//去队首
head++;//出队
for(int i=1; i<=6; i++)//遍历六种情况
{
if(i==1)//装满1
{
to.k1=a;
to.k2=now.k2;
}
else if(i==2)//装满2
{
to.k1=now.k1;
to.k2=b;
}
else if(i==3)//倒掉1
{
to.k1=0;
to.k2=now.k2;
}
else if(i==4)//倒掉2
{
to.k1=now.k1;
to.k2=0;
}
else if(i==5)//1->2
{
if(now.k1+now.k2<=b)//如果装不满b
{
to.k1=0;
to.k2=now.k1+now.k2;
}
else
{
to.k1=now.k1+now.k2-b;
to.k2=b;
}
}
else if(i==6)//2->1
{
if(now.k1+now.k2<=a)//如果装不满a
{
to.k1=now.k1+now.k2;
to.k2=0;
}
else
{
to.k1=a;
to.k2=now.k1+now.k2-a;
}
}
to.op=i;//记录操作
if(!vis[to.k1][to.k2])//当前状态没有被标记
{
vis[to.k1][to.k2]=1;//标记
to.step=now.step+1;//下一步
to.pre=head-1;//记录下标
q[tail].k1=to.k1;
q[tail].k2=to.k2;
q[tail].op=to.op;
q[tail].step=to.step;
q[tail].pre=to.pre;
tail++;
//以上为入队操作
if(to.k1==c||to.k2==c)//满足条件
{
flag=1;//状态标记为1
step=to.step;//最终的步数
lastindex=tail-1;//最终的标号
return;
}
}
}
}
}
int main()
{
while(~scanf("%d%d%d",&a,&b,&c))
{
flag=0;
step=0;
bfs();
if(flag)
{
printf("%d\n",step);
id[step]=lastindex;
for(int i=step-1; i>=1; i--)
{
id[i]=q[id[i+1]].pre;
}
for(int i=1; i<=step; i++)//判断下标并输出结果
{
if(q[id[i]].op==1)printf("FILL(1)\n");
if(q[id[i]].op==2)printf("FILL(2)\n");
if(q[id[i]].op==3)printf("DROP(1)\n");
if(q[id[i]].op==4)printf("DROP(2)\n");
if(q[id[i]].op==5)printf("POUR(1,2)\n");
if(q[id[i]].op==6)printf("POUR(2,1)\n");
}
}
else
printf("impossible\n");
}
return 0;
}
和非常可乐哪道题有很相似的地方\
