波松分酒问题 C++求最优解.

/*
请设计程序解决“波松分酒问题”
问题如下:

某人有12品脱啤酒一瓶,想从中倒出6品脱,但他没有6品脱的容器,
仅有一个8品脱和一个5品脱的容器,怎样才能将啤酒分为两个6品脱?

抽象分析:

b = 大容器,也表示容积
s = 小容器,也表示容积
(f),(h),(e) 状态f=满, e=空, h=数字,表示容量

运算一: b(f) - s(e)  =>  b(b - s), s(f)
变例    b(h) - s(e)  =>  b(h - s), s(f)

运算二: b(e) + s(f)  =>  b(s), s(e)
变例    b(h) + s(f)  =>  b(f), s(s - b + h)

引出    b(f) - s(h)
        b(h) - s(h)

        b(e) + s(h)
        b(h) + s(h)

如果以瓶中酒的数量为节点, 通过一次以上运算可达到节点之间认为连通.
此题可转化为一个有向图的搜索问题.
即找出.指定节点(12, 0, 0) 和 (6, 6, 0)之间的最小路径.

*/
#include <cstdio>
#include <deque>
#include <map>
#include <utility>
#include <queue>

static int big_max_value[] =
{
    12, 8, 12
};
static int small_max_value[] =
{
    8, 5, 5
};
static const int big_offset[] =
{
    0, 1, 0
};
static const int small_offset[] =
{
    1, 2, 2
};


//节点定义
class Node
{
    unsigned char mBig;
    unsigned char mMid;
    unsigned char mSmall;

public:
    static void InitMaxValue(int max1, int max2, int max3)
    {
        big_max_value[0] = max1;
        big_max_value[1] = max2;
        big_max_value[2] = max1;

        small_max_value[0] = max2;
        small_max_value[1] = max3;
        small_max_value[2] = max3;
    }

    Node() : mBig(0), mMid(0), mSmall(0)
    {
    }

    Node(unsigned char a, unsigned char b, unsigned char c) : mBig(a), mMid(b), mSmall(c)
    {
    }


    enum OPCODE
    {
        BIG_OP_MIDDLE               = 0,
        MIDDLE_OP_SMALL,
        BIG_OP_SMALL,
        OP_LAST
    };


    //减运算
    void sub(OPCODE op)
    {
        int big_max = big_max_value[op];
        int small_max = small_max_value[op];

        char& big = *(reinterpret_cast<char*>(this) + big_offset[op]);
        char& small = *(reinterpret_cast<char*>(this) + small_offset[op]);

        if (big > (small_max - small))
        {
            big -= (small_max - small);
            small = small_max;
        }
        else
        {
            small += big;
            big = 0;
        }
    }

    //加运算
    void add(OPCODE op)
    {
        int big_max = big_max_value[op];
        int small_max = small_max_value[op];

        char& big = *(reinterpret_cast<char*>(this) + big_offset[op]);
        char& small = *(reinterpret_cast<char*>(this) + small_offset[op]);

        if (small > big_max - big)
        {
            small -= big_max - big;
            big = big_max;
        }
        else
        {
            big += small;
            small = 0;
        }
    }

    bool check(int value)
    {
        if (mBig == value || mMid == value || mSmall == value)
        {
            return true;
        }
        return false;
    }

    void print() const
    {
        printf("status [%d]=%2d, [%d]=%2d, [%d]=%2dn", big_max_value[0], mBig, big_max_value[1], mMid,
            small_max_value[2], mSmall);
    }

    //相等性判定
    friend bool operator==(Node const & a, Node const & b)
    {
        return memcmp(&a, &b, sizeof(Node)) == 0;
    }

    friend bool operator <(Node const & a, Node const & b)
    {
        return memcmp(&a, &b, sizeof(Node)) < 0;
    }
};


template <class T>
void Search(T start, int value)
{
    typedef std::pair<T, T> NodeValueType;

    typedef std::map<T, T> NodeSet;
    typedef NodeSet::iterator NodeSetIter;

    typedef std::queue<NodeValueType, std::deque<NodeValueType> > NodeQueue;

    NodeSet visited;
    NodeQueue searchQueue;
    NodeValueType last;

    searchQueue.push(std::make_pair(start, start));

    while (!searchQueue.empty())
    {
        NodeValueType cur = searchQueue.front();
        searchQueue.pop();

        visited.insert(cur);
        if (cur.first.check(value))
        {
            last = cur;
            break;
        }

        for (int i = 0; i < Node::OP_LAST; i++)
        {
            Node next1 = cur.first;
            next1.sub(static_cast<Node::OPCODE>(i));

            if (visited.find(next1) == visited.end())
            {
                searchQueue.push(std::make_pair(next1, cur.first));
            }

            Node next2 = cur.first;
            next2.add(static_cast<Node::OPCODE>(i));

            if (visited.find(next2) == visited.end())
            {
                searchQueue.push(std::make_pair(next2, cur.first));
            }
        }
    }

    NodeSetIter cur = visited.find(last.first);

    while (!(cur->first == start))
    {
        cur->first.print();
        cur = visited.find(cur->second);
    }
    cur->first.print();
}

int main()
{
    puts("某人有12品脱啤酒一瓶,想从中倒出6品脱,但他没有6品脱的容器,n"
         "仅有一个8品脱和一个5品脱的容器,怎样才能将啤酒分为两个6品脱?n");

    for (int i = 0; i < 12; i++)
    {
        printf("---查找取得%d品脱的最少步骤,逆序------------n", i);
        Search(Node(12, 0, 0), i);
    }

    puts("再解一个由13品脱啤酒,却一个9品脱和一个5品脱的容器n");

    Node::InitMaxValue(13, 9, 5);
    for (int i = 0; i < 12; i++)
    {
        printf("---查找取得%d品脱的最少步骤,逆序------------n", i);
        Search(Node(13, 0, 0), i);
    }
    return 0;
}

实际上的最后一步,结果应是(6,6,0)但事实上我只做到出现一个6的情况.原因是并非所有结果都有两个相同的值.以下是我做出来的12,8,5的最优解法:
某人有12品脱啤酒一瓶,想从中倒出6品脱,但他没有6品脱的容器,
仅有一个8品脱和一个5品脱的容器,怎样才能将啤酒分为两个6品脱?

---查找取得0品脱的最少步骤,逆序------------
status [12]=12, [8]= 0, [5]= 0
---查找取得1品脱的最少步骤,逆序------------
status [12]= 1, [8]= 8, [5]= 3
status [12]= 9, [8]= 0, [5]= 3
status [12]= 9, [8]= 3, [5]= 0
status [12]= 4, [8]= 3, [5]= 5
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
---查找取得2品脱的最少步骤,逆序------------
status [12]= 2, [8]= 5, [5]= 5
status [12]= 7, [8]= 5, [5]= 0
status [12]= 7, [8]= 0, [5]= 5
status [12]=12, [8]= 0, [5]= 0
---查找取得3品脱的最少步骤,逆序------------
status [12]= 4, [8]= 3, [5]= 5
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
---查找取得4品脱的最少步骤,逆序------------
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
---查找取得5品脱的最少步骤,逆序------------
status [12]= 7, [8]= 0, [5]= 5
status [12]=12, [8]= 0, [5]= 0
---查找取得6品脱的最少步骤,逆序------------
status [12]= 1, [8]= 6, [5]= 5
status [12]= 1, [8]= 8, [5]= 3
status [12]= 9, [8]= 0, [5]= 3
status [12]= 9, [8]= 3, [5]= 0
status [12]= 4, [8]= 3, [5]= 5
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
---查找取得7品脱的最少步骤,逆序------------
status [12]= 7, [8]= 0, [5]= 5
status [12]=12, [8]= 0, [5]= 0
---查找取得8品脱的最少步骤,逆序------------
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
---查找取得9品脱的最少步骤,逆序------------
status [12]= 9, [8]= 3, [5]= 0
status [12]= 4, [8]= 3, [5]= 5
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
---查找取得10品脱的最少步骤,逆序------------
status [12]=10, [8]= 2, [5]= 0
status [12]= 5, [8]= 2, [5]= 5
status [12]= 5, [8]= 7, [5]= 0
status [12]= 0, [8]= 7, [5]= 5
status [12]= 7, [8]= 0, [5]= 5
status [12]=12, [8]= 0, [5]= 0
---查找取得11品脱的最少步骤,逆序------------
status [12]=11, [8]= 0, [5]= 1
status [12]= 3, [8]= 8, [5]= 1
status [12]= 3, [8]= 4, [5]= 5
status [12]= 8, [8]= 4, [5]= 0
status [12]= 8, [8]= 0, [5]= 4
status [12]= 0, [8]= 8, [5]= 4
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
注意这个解法通用性很强,还可以解其它的组合:如最后的13,9,5.

posted on 2004-08-12 13:14  Anthony-黄亮  阅读(198)  评论(0编辑  收藏  举报

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