[kuangbin带你飞]专题一 简单搜索H - Pots(POJ 3414)

H - Pots

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 3414

Appoint description:  System Crawler  (2015-11-09)

Description

You 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.

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：有两个杯子容量是A和B，水是无限的。得到C的水。

直接bfs

以两个杯中的水为状态，记得判重。

特判的数据要好好看看是否对，我就错了一发，搞了半天还不知道错在哪？

#include
  
   
#include
   
    
#include
    
     
#include
     
      
using namespace std;

char str[7][10] = {" ", "FILL(2)", "FILL(1)", "POUR(2,1)", "POUR(1,2)", "DROP(2)", "DROP(1)"};

struct node
{
    int a, b;//状态
    node() {}
    node(int aa, int bb)
    {
        a = aa;
        b = bb;
    }

};

struct
{
    int dir;//方式
    int next;//上一个状态
}vis[100000];//判重

queue
      
       que;

int ans;
void prin(int n)//输出路径
{
    if(n == 0)
        printf("%d\n", ans) ;
    else
    {
        ans++;
        prin(vis[n].next);
        printf("%s\n", str[ vis[n].dir ] );
    }
}

int a, b, c;
int f(int x, int y, node tmp, int n)
{

    if(!vis[x*200+y].dir)
    {
//        printf("\t%d==%d--%d\n", n, x, y);
        vis[x*200+y].next = tmp.a*200+tmp.b;//记录上次的位置，便于输出路径
        vis[x*200+y].dir = n;   //记录方式
        if(x == c || y == c) return x*200+y;//如果满足条件返回
        que.push(node(x, y));
    }
    return 0;
}

int bfs()
{
    while(!que.empty()) que.pop();
    que.push(node(0, 0));
    memset(vis, 0, sizeof(vis));
    vis[0].dir = -1;
    vis[0].next = -1;
    while(!que.empty())
    {
        node tmp = que.front();
        que.pop();
        int x = tmp.a, y = b;//把B倒满
        int temp = f(x, y, tmp, 1); if(temp) return temp;

        x = a, y = tmp.b;//把A倒满
        temp = f(x, y, tmp, 2); if(temp) return temp;

        x = tmp.a+tmp.b;//B中的水倒入A中
        if(x > a) x = a, y = tmp.a+tmp.b-a;
        else y = 0;
        temp = f(x, y, tmp, 3); if(temp) return temp;

        y = tmp.a+tmp.b;//A中的水倒入B中
        if(y > b) y = b, x = tmp.a+tmp.b-b;
        else x = 0;
        temp = f(x, y, tmp, 4); if(temp) return temp;

        x = tmp.a, y = 0;//把B中的水倒掉
        temp = f(x, y, tmp, 5); if(temp) return temp;

        x = 0, y = tmp.b;//把A中的水倒掉
        temp = f(x, y, tmp, 6); if(temp) return temp;
    }
    return -100;
}

int main(void)
{
    while(~scanf("%d%d%d", &a, &b, &c))
    {
        ans = 0;
        if(0 == c)//
        {
            printf("0\n");
            continue;
        }
        int tmp = bfs();
        if(tmp == -100) printf("impossible\n");
        else
        {
            prin(tmp);
        }
    }
    return 0;
}