abc.cpp

分酒问题C++源代码

#include <stdio>
#include <deque>
#include <map>
#include <utility>
#include <queue>

static int big_max_value[] =
{
    8, 5, 8
};
static int small_max_value[] =
{
    5, 3, 3
};
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("已知有三个容量分别为3千克,5千克和8千克的并且是没有刻度的酒瓶,n"
         "3千克和5千克的瓶子均装满了酒,而8千克的瓶子为空,现要求仅用"
         "这三个酒瓶将这些酒均分为两个4千克并分别装入5千克和8千克的瓶子中.n");

    for (int i = 0; i < 8; i++)
    {
        printf("---查找取得%d千克的最少步骤,逆序------------n", i);
        Search(Node(8, 0, 0), i);
    }

    return 0;
}