POJ-1606 Jugs

Jugs

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4076		Accepted: 2411		Special Judge

Description

In the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fill the 5-gallon jug with exactly 4 gallons. This problem generalizes that puzzle. 

You have two jugs, A and B, and an infinite supply of water. There are three types of actions that you can use: (1) you can fill a jug, (2) you can empty a jug, and (3) you can pour from one jug to the other. Pouring from one jug to the other stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons and B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A. 

A problem is given by a triple (Ca,Cb,N), where Ca and Cb are the capacities of the jugs A and B, respectively, and N is the goal. A solution is a sequence of steps that leaves exactly N gallons in jug B. The possible steps are 

fill A 
fill B 
empty A 
empty B 
pour A B 
pour B A 
success 

where "pour A B" means "pour the contents of jug A into jug B", and "success" means that the goal has been accomplished. 

You may assume that the input you are given does have a solution.

Input

Input to your program consists of a series of input lines each defining one puzzle. Input for each puzzle is a single line of three positive integers: Ca, Cb, and N. Ca and Cb are the capacities of jugs A and B, and N is the goal. You can assume 0 < Ca <= Cb and N <= Cb <=1000 and that A and B are relatively prime to one another.

Output

Output from your program will consist of a series of instructions from the list of the potential output lines which will result in either of the jugs containing exactly N gallons of water. The last line of output for each puzzle should be the line "success". Output lines start in column 1 and there should be no empty lines nor any trailing spaces.

Sample Input

3 5 4 
5 7 3 

Sample Output

fill B 
pour B A 
empty A 
pour B A 
fill B 
pour B A 
success 
fill A 
pour A B 
fill A 
pour A B 
empty B 
pour A B 
success 

/*注意：本題是讓 b 中達到目標狀態   用到隊列 + 記憶化搜索
//代碼一：純模擬---從討論區看見的，由於本題沒有要求輸出的結果是最少的操作次數，所以這種方法也是可以的
#include <iostream>

using namespace std;

int main()
{
    int ca,cb,n;
    while (cin >> ca >> cb >> n)
    {
        int bnow;
        int b = 0;
        while (b != n)
        {
            for (int i = 0;i <= (cb - b) / ca;i++)
            {
                cout << "fill A" << endl;
                cout << "pour A B" << endl;
                bnow = b + ca * (i + 1);
                if (bnow == n)
                {
                    break;
                }
            }
            if (bnow == n)
            {
                break;
            }
            cout << "empty B" << endl;
            int a;
            a = ca - (cb - b) % ca;
            cout << "pour A B" << endl;
            b = a;
            if (b == n)
            {
                break;
            }
        }
        cout << "success" << endl;
    }
    return 0;
}
*/
/*
//代碼二： 用STL隊列，可以保證結果肯定是最少的操作次數-----搜索路徑算法還未完成
#include <iostream>
#include <cstring>
//#include <queue>

using namespace std;

struct state
{
    int a;
    int b;
    int opt;
    int last_opt;
    int tot_opt;
};
char cmd[6][10] = {"fill A", "fill B", "empty A", "empty B", "pour A B", "pour B A"};
bool visit[1001][1001];
int path[200];
int ca, cb, n;


int BFS(int sa, int sb, int steps)
{
    queue <state> q;
    front = tail = 0;
    state t1, t2;
    t1.a = 0;
    t1.b = 0;
    t1.opt = -1;
    t1.last_opt = -1;
    t1.tot_opt = 0;
    q.push(t1);
    while(!q.empty())
    {
        t2 = q.front();
        q.pop();
        if(t2.b ==n)
        {
            。。。。。。。。
            。。。。。。。。
          //  因為用的是stl中的queue，所以這里還沒想到應該咋實現才能回溯遍歷所有的路徑
            。。。。。。。。
        }
        else
        {
            state t3;
            if(t2.a != ca && !visit[ca][t2.b])  //fill A
            {
                t3.a = ca;
                t3.b = t2.b;
                t3.opt = 0;
                t3.last_opt = t2.opt;
                t3.tot_opt = t2.tot_opt + 1;
                visit[ca][t2.b] = true;
                q.push(t3);
            }
            if(t2.b != cb && !visit[t2.a][cb])   // fill B
            {
                t3.a = t2.a;
                t3.b = cb;
                t3.opt = 1;
                t3.last_opt = t2.opt;
                t3.tot_opt = t2.tot_opt + 1;
                visit[t2.a][cb] = true;
                q.push(t3);
            }
            if(t2.a != 0 && !visit[0][t2.b])    //  empty A
            {
                t3.a = 0;
                t3.b = t2.b;
                t3.opt = 2;
                t3.last_opt = t2.opt;
                t3.tot_opt = t2.tot_opt + 1;
                visit[0][t2.b] = true;
                q.push(t3);
            }
            if(t2.b != 0 && !visit[t2.a][0])   // empty B
            {
                t3.a = t2.a;
                t3.b = 0;
                t3.opt = 3;
                t3.last_opt = t2.opt;
                t3.tot_opt = t2.tot_opt + 1;
                visit[t2.a][0] = true;
                q.push(t3);
            }
            if(sb != cb && t2.a != 0)   // pour A to B
            {
                if(t2.a <= cb - t2.b)
                {
                    t3.a = 0;
                    t3.b = t2.b + t2.a;
                }
                else
                {
                    t3.a = t2.a - (cb - t2.b);
                    t3.b = cb;
                }
                t3.opt = 4;
                t3.last_opt = t2.opt;
                t3.tot_opt = t2.tot_opt + 1;
                visit[t3.a][t3.b] = true;
                q.push(t3);
            }
            if(t2.a != ca && cb != 0)    //  pour B to A
            {
                if(t2.b <= ca - t2.a)
                {
                    t3.b = 0;
                    t3.a = t2.a + t2.b;
                }
                else
                {
                    t3.a = ca;
                    t3.b = t2.b - (ca - t2.a);
                }
                t3.opt = 5;
                t3.last_opt = t2.opt;
                t3.tot_opt = t2.tot_opt + 1;
                visit[t3.a][t3.b] = true;
                q.push(t3);
            }
        }
    }
}

int main()
{
    while(cin >> ca >> cb >> n)
    {
        memset(visit, false, sizeof(visit));
        memset(path, 0, sizeof(path));
        visit[0][0] = true;
        int tot_step = BFS(0, 0, 0);
        for(int i = 1; i <= tot_step; ++i)
            cout << cmd[path[i]] << endl;
        cout << "success" <<endl;
    }
    return 0;
}
*/
//代碼三：------AC
#include <iostream>
#include <cstring>

using namespace std;

struct state
{
    int a;   //記錄當前a中的量
    int b;   //記錄當前b中的量
    int opt;  //記錄當前狀態是由前一狀態怎樣得來的，即記錄的是操作數----對應的是cmd數組的一維下標
    int last_path;  //記錄上一步所在狀態的隊列下標，方便回溯查找路徑
    int tot_opt;    //記錄到底本狀態總共用了多少步
}q[10000];
char cmd[6][10] = {"fill A", "fill B", "empty A", "empty B", "pour A B", "pour B A"};
bool visit[1001][1001];
int path[200];
int ca, cb, n;


int BFS(int sa, int sb, int steps)
{
    int front, tail;
    front = tail = 0;
    state t1, t2;
    t1.a = 0;
    t1.b = 0;
    t1.opt = -1;
    t1.last_path = -1;
    t1.tot_opt = 0;
    q[tail++] = t1;
    while(front != tail)
    {
        t2 = q[front++];
        if(t2.b == n)
        {
            state t = t2;
            for(int i = t.tot_opt; i > 0; --i)
            {
               path[i] = t.opt;
               t = q[t.last_path];
            }
            return t2.tot_opt;
        }
        else
        {
            state t3;
            if(t2.a != ca && !visit[ca][t2.b])  //fill A
            {
                t3.a = ca;
                t3.b = t2.b;
                t3.opt = 0;
                t3.last_path = front - 1;
                t3.tot_opt = t2.tot_opt + 1;
                visit[ca][t2.b] = true;
                q[tail++] = t3;
            }
            if(t2.b != cb && !visit[t2.a][cb])   // fill B
            {
                t3.a = t2.a;
                t3.b = cb;
                t3.opt = 1;
                t3.last_path = front - 1;
                t3.tot_opt = t2.tot_opt + 1;
                visit[t2.a][cb] = true;
                q[tail++] = t3;
            }
            if(t2.a != 0 && !visit[0][t2.b])    //  empty A
            {
                t3.a = 0;
                t3.b = t2.b;
                t3.opt = 2;
                t3.last_path = front - 1;
                t3.tot_opt = t2.tot_opt + 1;
                visit[0][t2.b] = true;
                q[tail++] = t3;
            }
            if(t2.b != 0 && !visit[t2.a][0])   // empty B
            {
                t3.a = t2.a;
                t3.b = 0;
                t3.opt = 3;
                t3.last_path = front - 1;
                t3.tot_opt = t2.tot_opt + 1;
                visit[t2.a][0] = true;
                q[tail++] = t3;;
            }
            if(sb != cb && t2.a != 0)   // pour A to B
            {
                if(t2.a <= cb - t2.b)
                {
                    t3.a = 0;
                    t3.b = t2.b + t2.a;
                }
                else
                {
                    t3.a = t2.a - (cb - t2.b);
                    t3.b = cb;
                }
                t3.opt = 4;
                t3.last_path = front - 1;
                t3.tot_opt = t2.tot_opt + 1;
                visit[t3.a][t3.b] = true;
                q[tail++] = t3;
            }
            if(t2.a != ca && cb != 0)    //  pour B to A
            {
                if(t2.b <= ca - t2.a)
                {
                    t3.b = 0;
                    t3.a = t2.a + t2.b;
                }
                else
                {
                    t3.a = ca;
                    t3.b = t2.b - (ca - t2.a);
                }
                t3.opt = 5;
                t3.last_path = front - 1;
                t3.tot_opt = t2.tot_opt + 1;
                visit[t3.a][t3.b] = true;
                q[tail++] = t3;
            }
        }
    }
}

int main()
{
    while(cin >> ca >> cb >> n)
    {
        memset(visit, false, sizeof(visit));
        memset(path, 0, sizeof(path));
        visit[0][0] = true;
        int tot_step = BFS(0, 0, 0);
        for(int i = 1; i <= tot_step; ++i)
            cout << cmd[path[i]] << endl;
        cout << "success" <<endl;
    }
    return 0;
}