/*
請設計程序解決“波松分酒問題”
問題如下：
某人有12品脫啤酒一瓶，想從中倒出6品脫，但他沒有6品脫的容器，
僅有一個8品脫和一個5品脫的容器，怎樣才能將啤酒分爲兩個6品脫？
抽象分析:
b = 大容器,也表示容積
s = 小容器,也表示容積
(f),(h),(e) 狀態f=滿, e=空, h=數字,表示容量
運算一: b(f) - s(e) => b(b - s), s(f)
變例 b(h) - s(e) => b(h - s), s(f)
運算二: b(e) + s(f) => b(s), s(e)
變例 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)
如果以瓶中酒的數量爲節點, 通過一次以上運算可達到節點之間認爲連通.
此題可轉化爲一個有向圖的搜索問題.
即找出.指定節點(12, 0, 0) 和 (6, 6, 0)之間的最小路徑.
*/
#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
};
//節點定義
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
};
//減運算
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;
}
}
//加運算
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品脫啤酒一瓶，想從中倒出6品脫，但他沒有6品脫的容器，n"
"僅有一個8品脫和一個5品脫的容器，怎樣才能將啤酒分爲兩個6品脫？n");
for (int i = 0; i < 12; i++)
{
printf("---查找取得%d品脫的最少步驟,逆序------------n", i);
Search(Node(12, 0, 0), i);
}
puts("再解一個由13品脫啤酒，卻一個9品脫和一個5品脫的容器n");
Node::InitMaxValue(13, 9, 5);
for (int i = 0; i < 12; i++)
{
printf("---查找取得%d品脫的最少步驟,逆序------------n", i);
Search(Node(13, 0, 0), i);
}
return 0;
}
實際上的最後一步,結果應是(6,6,0)但事實上我只做到出現一個6的情況.原因是並非所有結果都有兩個相同的值.以下是我做出來的12,8,5的最優解法:
某人有12品脫啤酒一瓶，想從中倒出6品脫，但他沒有6品脫的容器，
僅有一個8品脫和一個5品脫的容器，怎樣才能將啤酒分爲兩個6品脫？
---查找取得0品脫的最少步驟,逆序------------
status [12]=12, [8]= 0, [5]= 0
---查找取得1品脫的最少步驟,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品脫的最少步驟,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品脫的最少步驟,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品脫的最少步驟,逆序------------
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
---查找取得5品脫的最少步驟,逆序------------
status [12]= 7, [8]= 0, [5]= 5
status [12]=12, [8]= 0, [5]= 0
---查找取得6品脫的最少步驟,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品脫的最少步驟,逆序------------
status [12]= 7, [8]= 0, [5]= 5
status [12]=12, [8]= 0, [5]= 0
---查找取得8品脫的最少步驟,逆序------------
status [12]= 4, [8]= 8, [5]= 0
status [12]=12, [8]= 0, [5]= 0
---查找取得9品脫的最少步驟,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品脫的最少步驟,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品脫的最少步驟,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
注意這個解法通用性很強,還可以解其它的組合:如最後的13,9,5.