ZOJ 1005 Jugs(模拟 special judge,也可以BFS)

Jugs Time Limit: 2 Seconds       Memory Limit: 65536 KB       Special Judge

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

很经典的用两个杯子，容量分别为ca,cb，然后只能使用装满，清空，或者pour A B ，pour B A 。

   由于两个数是互质的，那么ca==cb也是不可能的。。每次A空了，就装满，然后往B里面倒，B满了就清空，然后再由A向B中倒。反正我是找了几对互质的数，发现所有的数字都是可以找到的1~cb。

AC代码：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

int main()
{
    int ca,cb,x;
    int a,b,t;
    while(cin>>ca>>cb>>x)
    {
        a=0,b=0;
        while(1)
        {
            if(a==0)     //A装满
            {
                puts("fill A");
                a=ca;
            }
            if(a==x||b==x) break;
            puts("pour A B");
            if(b+a>=cb)     //B装满了
                a=b+a-cb,b=cb;
            else
               b=b+a,a=0;
            if(a==x||b==x) break;
            if(b==cb)
            {
                puts("empty B");
                b=0;
            }
        }
        puts("success");
    }
    return 0;
}

/*
3 5 4
5 7 3
7 7 2
*/

不过这个题目要求最小的步骤的话，就必须使用BFS了。