【算法】算法之美—Jugs

题目概述：Jugs

   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:

• you can fill a jug,

• you can empty a jug

• 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

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简单描述

   题目意思仍是比较简单，简单概述一下就是：有无限的水和两个水桶，一个容量 ca升，一个容量 cb升，问怎么用两个桶获得 n升水，要求比较简单，就不复述了。

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题目分析

   简单分析下此题目：

1. 两个水桶 A和B，则须要两个变量来表示其容量，设为 ca和cb

2. 目标变量 n

3. 水在倒的过程当中，两个水桶的当前水量会发生变化，须要两个变量表示当前量，设为 x和y

4. 目标量可能由 A获得，也可能由 B获得，因此须要对 A和B进行判断  
   

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解题算法

   注释丰富，就很少说了

#include < stdio.h>
int main()
{
    int ca,cb,n,x,y;
    /* Ca, Cb, and N. Ca and Cb are the capacities of jugs A and B, and N is the goal.
     * 即：ca为 A的容量，cb为 B的容量，N为要获得的水
     * x为 ca当前已有的水
     * y为 cb当前已有的水
     */
    while( scanf("%d %d %d",&ca,&cb,&n)!=EOF)
    {
        x=y=0;
        while(1)
        {
            printf("fill A\n");
            x=ca;
            if(x==n)    /* 若是 A获得目标 */
            {
                printf("success\n");
                break;  /* 成功，退出循环 */
            }
            while(x>0)
            {
                if((cb-y)>=x)    /* B可以容纳的水 > A当前的水 */
                {
                    printf("pour A B\n");   /* 则将 A中的水所有倒入 B中 */
                    y=y+x;  /* y增长 x升水 */
                    x=0;
                }
                else
                {
                    printf("pour A B\n");
                    x=x-(cb-y); /* A中还剩有水 */
                    y=cb;   /* B中装满 */
                }
                if(x==n)    /* A获得目标 */
                {
                    printf("success\n");
                    break;
                }
                if(y==n)    /* B获得目标 */
                    break;
                if(y==cb)   /* B空 */
                {
                    printf("empty B\n");
                    y=0;
                }
            }
            if(x==n)
                break;
            if(y==n)
            {
                printf("success\n");
                break;
            }
        }
    }
    return 0;
}