Mother's Milk（DFS）

最新推荐文章于 2018-08-04 11:10:10 发布

原创最新推荐文章于 2018-08-04 11:10:10 发布 · 223 阅读

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本文介绍了一个经典的牛奶倒换问题，通过使用深度优先搜索（DFS）算法，探讨了如何在不同容量的桶之间倒换牛奶以达到特定目标状态的方法。文章提供了一段C++代码示例，展示了如何有效地解决问题，并通过实例说明了可能得到的结果。

Mother's Milk

Farmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk. Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.

Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.

PROGRAM NAME: milk3

INPUT FORMAT

A single line with the three integers A, B, and C.

SAMPLE INPUT (file milk3.in)

8 9 10

OUTPUT FORMAT

A single line with a sorted list of all the possible amounts of milk that can be in bucket C when bucket A is empty.

SAMPLE OUTPUT (file milk3.out)

1 2 8 9 10

SAMPLE INPUT (file milk3.in)

2 5 10

SAMPLE OUTPUT (file milk3.out)

5 6 7 8 9 10

题意：

给A,B,C三个瓶子的容积，一开始为0,0,C（C为满）的状态。随后互相可以倒牛奶，每次倒都是一次倒完，或者倒满容器为止。求当A=0时，C牛奶容积的可能性。从小到大输出结果。

思路：

dfs。标记六种倒的过程，分别为A倒B，A倒C，B倒A，B倒C，C倒A，C倒B。用vis数组标记每种状态是否曾经到达过。最后将A==0时候C的数存到num数组中，最后输出vis数组即可。

AC：

/*  
TASK:milk3  
LANG:C++  
ID:sum-g1  
*/ 
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int res[25],ans,a,b,c,vis[20][20][20];

void dfs(int na,int nb,int nc)
{
    if(vis[na][nb][nc]) return;
    vis[na][nb][nc]=1;   //标记状态
    if(!na)   //判断不是!na&&nc,nc可以为0
    {
        ans++;
        res[ans]=nc;
    }
    if(na&&nb!=b)
    {
        if(b-nb>=na)  dfs(0,na+nb,nc);
        else          dfs(na+nb-b,b,nc);
    }
    if(na&&nc!=c)
    {
        if(c-nc>=na)  dfs(0,nb,na+nc);
        else          dfs(na+nc-c,nb,c);
    }
    if(nb&&na!=a)
    {
        if(a-na>=nb)  dfs(na+nb,0,nc);
        else          dfs(a,nb+na-a,nc);
    }
    if(nb&&nc!=c)
    {
        if(c-nc>=nb)  dfs(na,0,nc+nb);
        else          dfs(na,nb+nc-c,c);
    }
    if(nc&&na!=a)
    {
        if(a-na>=nc)  dfs(na+nc,nb,0);
        else          dfs(a,nb,nc+na-a);
    }
    if(nc&&nb!=b)
    {
        if(b-nb>=nc)  dfs(na,nb+nc,0);
        else          dfs(na,b,nc+nb-b);
    }
}

int main()
{
    freopen("milk3.in","r",stdin);    
    freopen("milk3.out","w",stdout);  
    ans=0;
    memset(res,0,sizeof(res));
    memset(vis,0,sizeof(vis));
    scanf("%d%d%d",&a,&b,&c);
    dfs(0,0,c);
    sort(res+1,res+1+ans);
    for(int i=1;i<=ans;i++)
    {
        printf("%d",res[i]);
        i==ans?printf("\n"):printf(" ");
    }
    return 0;
}

总结：

1.应该标记的是a,b,c的状态，而不是标记曾经走过哪个过程，标记过程会导致重复搜索，比如a->b，b->a则会返回去原来的状态；

2.c的状态可能为0；