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  • 波松分酒问题 C++求最优解. Anthony

    /*
    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.

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  • 原文地址:https://www.cnblogs.com/ahuangliang/p/5309273.html
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