poj 3414 (POTS) (BFS)

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.

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).

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’.

/**解题思路：
就是一个简单的BFS,仔细搞搞就行了

**/


/**
2015 - 09 - 16 下午
Author: ITAK

Motto:

今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 105;
const double eps = 1e-7;
bool vis[maxn][maxn];
const int dir[4][2]= {1, 0, 0, 1, -1, 0, 0, -1};
char map[maxn][maxn];

int a, b, k;
struct node
{
    int vola, volb, step;
    char str[maxn][maxn];
};

bool bfs()
{
    memset(vis, false, sizeof(vis));
    queue<node> que;
    node p, q;
    p.vola = 0, p.volb = 0, p.step = 0;
    que.push(p);
    vis[0][0] = 1;
    ///vis[p.vola][p.volb] = true;
    while(!que.empty())
    {
        p = que.front();
        que.pop();
        if(p.vola==k || p.volb == k)
        {
            cout<<p.step<<endl;
            for(int i=1; i<=p.step; i++)
                ///cout<<p.str[i]<<endl;
                printf("%s\n",p.str[i]);
            return true;
        }
        ///倒满 a
        if(p.vola == 0)
        {
            q = p;
            q.vola = a;
            q.step++;
            strcpy(q.str[q.step], "FILL(1)");
            if(!vis[q.vola][q.volb])
            {
                vis[q.vola][q.volb] = true;
                que.push(q);
            }
        }
        ///把 a 中的水倒出来
        else if(p.vola <= a)
        {
            q = p;
            q.vola = 0;
            q.step++;
            strcpy(q.str[q.step], "DROP(1)");
            if(!vis[q.vola][q.volb])
            {
                vis[q.vola][q.volb] = true;
                que.push(q);
            }
            ///a -> b
            if(p.volb < b)
            {
                q = p;
                if(q.volb + q.vola <= b)
                {
                    q.volb += q.vola;
                    q.vola = 0;
                }
                else
                {
                    q.vola = (q.vola+q.volb)-b;
                    q.volb = b;
                }
                q.step++;
                strcpy(q.str[q.step], "POUR(1,2)");
                if(!vis[q.vola][q.volb])
                {
                    vis[q.vola][q.volb] = true;
                    que.push(q);
                }
            }
        }
        ///把 b 倒满
        if(p.volb == 0)
        {
            q = p;
            q.volb = b;
            q.step++;
            strcpy(q.str[q.step], "FILL(2)");
            if(!vis[q.vola][q.volb])
            {
                vis[q.vola][q.volb] = true;
                que.push(q);
            }
        }
        ///把 b 中的水倒出来
        else if(p.volb <= b)
        {
            q = p;
            q.volb = 0;
            q.step++;
            strcpy(q.str[q.step],"DROP(2)");
            if(!vis[q.vola][q.volb])
            {
                vis[q.vola][q.volb] = true;
                que.push(q);
            }
            if(p.vola < a)
            {
                q = p;
                if(q.vola + q.volb <= a)
                {
                    q.vola += q.volb;
                    q.volb = 0;
                }
                else
                {
                    q.volb = (q.vola+q.volb)-a;
                    q.vola = a;
                }
                q.step++;
                strcpy(q.str[q.step], "POUR(2,1)");
                if(!vis[q.vola][q.volb])
                {
                    vis[q.vola][q.volb] = true;
                    que.push(q);
                }
            }
        }
    }
    return false;
}
int main()
{
    while(cin>>a>>b>>k)
    {
        bool ok = bfs();
        if(!ok)
            puts("impossible");
    }
    return 0;
}
/**
Sample Input
3 5 4
Sample Output
6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)
**/