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.