链接：
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 8253 | Accepted: 3499 | 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)
Source
算法：数组模拟bfs
题意：
思路：
code：
#include<stdio.h>
#include<string.h>
const int maxn = 110;
int vis[maxn][maxn]; //标记状态是否入队过
int a,b,c; //容器大小
int step; //最终的步数
int flag; //纪录是否能够成功
/* 状态纪录 */
struct Status{
int k1,k2; //当前水的状态
int op; //当前操作
int step; //纪录步数
int pre; //纪录前一步的下标
}q[maxn*maxn];
int id[maxn*maxn]; //纪录最终操作在队列中的编号
int lastIndex; //最后一个的编号
void bfs()
{
Status now, next;
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++;
memset(vis,0,sizeof(vis));
vis[0][0] = 1; //标记初始状态已入队
while(head < tail) //当队列非空
{
now = q[head]; //取出队首
head++; //弹出队首
if(now.k1 == c || now.k2 == c) //应该不会存在这样的情况, c=0
{
flag = 1;
step = now.step;
lastIndex = head-1; //纪录最后一步的编号
}
for(int i = 1; i <= 6; i++) //分别遍历 6 种情况
{
if(i == 1) //fill(1)
{
next.k1 = a;
next.k2 = now.k2;
}
else if(i == 2) //fill(2)
{
next.k1 = now.k1;
next.k2 = b;
}
else if(i == 3) //drop(1)
{
next.k1 = 0;
next.k2 = now.k2;
}
else if(i == 4) // drop(2);
{
next.k1 = now.k1;
next.k2 = 0;
}
else if(i == 5) //pour(1,2)
{
if(now.k1+now.k2 <= b) //如果不能够装满 b
{
next.k1 = 0;
next.k2 = now.k1+now.k2;
}
else //如果能够装满 b
{
next.k1 = now.k1+now.k2-b;
next.k2 = b;
}
}
else if(i == 6) // pour(2,1)
{
if(now.k1+now.k2 <= a) //如果不能够装满 a
{
next.k1 = now.k1+now.k2;
next.k2 = 0;
}
else //如果能够装满 b
{
next.k1 = a;
next.k2 = now.k1+now.k2-a;
}
}
next.op = i; //纪录操作
if(!vis[next.k1][next.k2]) //如果当前状态没有入队过
{
vis[next.k1][next.k2] = 1; //标记当前状态入队
next.step = now.step+1; //步数 +1
next.pre = head-1; //纪录前一步的编号
//q.push(next);
//q[tail] = next; 加入队尾
q[tail].k1 = next.k1; q[tail].k2 = next.k2;
q[tail].op = next.op; q[tail].step = next.step; q[tail].pre = next.pre;
tail++; //队尾延长
if(next.k1 == c || next.k2 == c) //如果达到目标状态
{
flag = 1; //标记成功
step = next.step; //纪录总步骤数
lastIndex = tail-1; //纪录最后一步在模拟数组中的编号
return;
}
}
}
}
}
int main()
{
while(scanf("%d%d%d", &a,&b,&c) != EOF)
{
flag = 0; //初始化不能成功
step = 0;
bfs();
if(flag)
{
printf("%d
", step);
id[step] = lastIndex; //最后一步在模拟数组中的编号
for(int i = step-1; i >= 1; i--)
{
id[i] = q[id[i+1]].pre; //向前找前一步骤在模拟数组中的编号
}
for(int i = 1; i <= step; i++)
{
if(q[id[i]].op == 1)
printf("FILL(1)
");
else if(q[id[i]].op == 2)
printf("FILL(2)
");
else if(q[id[i]].op == 3)
printf("DROP(1)
");
else if(q[id[i]].op == 4)
printf("DROP(2)
");
else if(q[id[i]].op == 5)
printf("POUR(1,2)
");
else if(q[id[i]].op == 6)
printf("POUR(2,1)
");
}
}
else printf("impossible
");
}
return 0;
}
小结：
其实这道题想来是大一第一学期最后一次学长上课,08级的黄超学长提到过的
两年了，一直到前几天居然都没有思路，昨天觉得还是要把基础的搜索练一下，又重新拿出来做，仔细分析了，题目还是很简单的，有时候还是过于浮躁。
做完后，1A 发现这题居然是在POJ 上直接 AC 的第一百题，两年了，今天才破百，博客量都快赶上刷题量了。实在是值得好好反思，有些时候很多东西真的得靠刷题量才能搞上去，所以这两年来，自己一直这么弱，现在只能做新生的题目,hdu 的多校完全沾不上边，也是有原因的。。。
然后和 kuangbin 大神吐槽了下, 他和我说，和他组队刷多校的有个人,两个月就刷了 2000+ 直跃 POJ 第八 http://poj.org/userstatus?user_id=neko13
虽然明显不科学，但是也证明了人家肯搞还是能做出很多题的，疯狂吧，再不疯狂我们都老了。