題目連結：

題目大意：

有兩個水杯，一開始都沒有裝水，給出兩個水杯的容量，和要達到的目標容量
每個水被有三種操作：１．裝滿水，２．倒掉所有水，３，將水倒入另一個杯子中．

思路：

兩個杯子，六種操作，，求最短路，並列印路徑，６入口bfs，跟非常可樂一題類似

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)

程式碼：

/*************************************************************************
        > File Name: poj_3414.cpp
      > Author: dulun
      > Mail: [email protected]
 

      > Created Time: 2016年04月13日 星期三 12時59分54秒
 ************************************************************************/

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#define LL long long
using namespace std;

const int N = 186;
struct Node
{
    int a, b, step;
    char str[N][N];
};

bool v[N][N];

void bfs(int a, int b, int c)
{
    memset(v, 0, sizeof(v));
    Node t, s;
    queue<Node> q;
    t = (Node){0, 0, 0};
    q.push(t);
    v[t.a][t.b] = true;

    while(!q.empty())
    {
        t = q.front();
        q.pop();
        v[t.a][t.b] = true;
        if(t.a == c || t.b == c)
        {
            printf("%d\n", t.step);
            for(int i = 1; i <= t.step; i++)
            {
                printf("%s\n", t.str[i]);
            }
            return ;
        }

        if(t.a == 0)
        {
            s = t;
            s.a = a;
            s.step++;
            strcpy(s.str[s.step], "FILL(1)");
            if(!v[s.a][s.b])
            {
                v[s.a][s.b] = true;
                q.push(s);
            }
        }
        else if(t.a <= a)
        {
            s = t;
            s.a = 0;
            s.step++;
            strcpy(s.str[s.step], "DROP(1)");
            if(!v[s.a][s.b])
            {
                v[s.a][s.b] = true;
                q.push(s);
            }
            if(t.b < b)
            {
                s = t;
                if(t.a+t.b <= b)
                {
                    s.a = 0;
                    s.b = t.a+t.b;
                }
                else{
                    s.a = t.a - (b-t.b);
                    s.b = b;
                }
                s.step++;
                strcpy(s.str[s.step], "POUR(1,2)");
                if(!v[s.a][s.b])
                {
                    v[s.a][s.b] = true;
                    q.push(s);
                }
            }
        }

        if(t.b == 0)
        {
            s = t;
            s.b = b;
            s.step++;
            strcpy(s.str[s.step], "FILL(2)");
            if(!v[s.a][s.b])
            {
                v[s.a][s.b] = true;
                q.push(s);
            }
        }
        else if(t.b <= b)
        {
            s = t;
            s.b = 0;
            s.step++;
            strcpy(s.str[s.step], "DROP(2)");
            if(!v[s.a][s.b])
            {
                v[s.a][s.b] = true;
                q.push(s);
            }
            if(t.a < a)
            {
                if(t.a+t.b < a)
                {
                    s = t;
                    s.a = t.a+t.b;
                    s.b = 0;
                }
                else{
                    s = t;
                    s.a = a;
                    s.b = t.b - (a-t.a);
                }
                s.step++;
                strcpy(s.str[s.step], "POUR(2,1)");
                if(!v[s.a][s.b])
                {
                    v[s.a][s.b] = true;
                    q.push(s);
                }
            }
        }

    }
    printf("impossible\n");
    return;

}

int main()
{
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);

    bfs(a, b, c);
    return 0;
}