（poj）3414 Pots （输出路径的广搜）

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 有两个杯子体积a和b升，有六个操作，经过几次操作可使至少其中一杯水有c升，并输出操作方法

方法 利用广搜，每种状态由可操作的状态向下搜索，找到一种就为操作次数最少的，找到输出操作方法

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include <math.h>
#include<queue>
#define ll long long
#define met(a,b) memset(a,b,sizeof(a));
#define N 405
using namespace std;
char str[6][50]={"FILL(2)","FILL(1)","POUR(2,1)","POUR(1,2)","DROP(1)","DROP(2)"};
int n,m,k,f;
int vis[N][N];
struct node
{
    int x,y,t;
    char s[500];///记录路径过程   s[i] i代表第i步做str[s[i]]这个操作
};
void dfs()
{
    queue<node> Q;
    node q,p;
    q.x=0;
    q.y=0;
    q.t=0;
    strcpy(q.s,"0");///记得初始化
    vis[0][0]=1;
    Q.push(q);
    while(Q.size())
    {
        p=Q.front();
        Q.pop();
        if(p.x==k || p.y==k)
        {
            f=1;
            printf("%d\n",p.t);
            for(int i=1; i<=p.t; i++)
                printf("%s\n",str[p.s[i]-‘0‘]);///根据p.s储存的输出路径
            break;
        }
        for(int i=0; i<6; i++)///每次一共就有六种操作可做，符合条件的才实行
        {
            q.t=p.t+1;
            q.x=0;
            q.y=0;
            strcpy(q.s,p.s);
            q.s[q.t]=i+‘0‘;
            if(i==0 && !p.y)///第二个杯子为空，装满
            {
                q.x=p.x;
                q.y=m;

            }
            else if(i==1 && !p.x)///第一个杯子为空，装满
            {
                q.x=n;
                q.y=p.y;
            }
            else if(i==2 && p.x<n&&p.y)///第一个杯子不满，第二个杯子不空，把第二个杯子的水倒入第一个杯子里
            {
                if(p.x+p.y<=n)
                {
                    q.x=p.x+p.y;
                    q.y=0;
                }
                else
                {
                    q.x=n;
                    q.y=p.x+p.y-n;
                }
            }
            else if(i==3 && p.y<m && p.x)///第二个杯子不满，第一个杯子不空，把第一个杯子的水倒入第二个杯子里
            {
                if(p.x+p.y<=m)
                {
                    q.x=0;
                    q.y=p.x+p.y;
                }
                else
                {
                    q.y=m;
                    q.x=p.x+p.y-m;

                }
            }
            else if(i==4 && p.x)///第一个杯子不空，倒完
            {
                q.x=0;
                q.y=p.y;
            }
            else if(i==5 && p.y)///第二个杯子不空，倒完
            {
                q.y=0;
                q.x=p.x;
            }
            if(!vis[q.x][q.y])///是否出现过，防止重复出现，死循环
            {
                vis[q.x][q.y]=1;
                Q.push(q);
            }
        }
    }
    return ;
}
int main()
{
    while(scanf("%d %d %d",&n,&m,&k)!=EOF)
    {
        met(vis,0);
        f=0;
        dfs();
        if(f==0)
            printf("impossible\n");
    }
    return 0;
}