UVa和POJ都有这道题。

不同的是UVa要求区分单复数，而POJ不要求。

使用STL做会比較简单，这里纯粹使用指针做了，很麻烦的指针操作，一不小心就错。

调试起来还是很费力的

本题理解起来也是挺费力的，要搞清楚怎样模拟也不easy啊，读题要非常细致。

纯指针的操作挺快的吧。

只是POJ 0ms，而UVa就0.2左右了。

三相链表：

1 仅仅要有叠起来的牌。那么就使用一个down指针指向以下的牌就能够了。

2 使用双向链表，能够方便前后遍历。

3 记得有了更新牌之后。又要又一次開始检查是否须要更新牌，这是模拟的须要。不然或WA的。

#include <stdio.h>

struct Node

{

	int size;

	Node *pre, *post;

	Node *down;

	char a, b;

	Node () : pre(NULL), post(NULL), down(NULL), size(1) {}

};

//insert n to m position

inline void insertNode(Node *&m, Node *&n)

{

	n->post = m->post;

	n->pre = m->pre;

	if (m->post) m->post->pre = n;//小心断链

	if (m->pre) m->pre->post = n;//小心断链

}

inline void takeoutNode(Node *&n)

{

	if (n->down)

	{

		Node *down = n->down;

		insertNode(n, down);

		return;

	}

	if (n->pre) n->pre->post = n->post;

	if (n->post) n->post->pre = n->pre;

}

inline void inStackNode(Node *&m, Node *&n)

{

	n->size = m->size+1;

	insertNode(m, n);

	n->down = m;

}

inline bool checkMovable(Node *n, Node *m)

{

	return n->a == m->a || n->b == m->b;

}

inline void pre3(Node *&n)

{

	if (n->pre) n = n->pre;

	if (n->pre) n = n->pre;

	if (n->pre) n = n->pre;

}

inline void pre1(Node *&n)

{

	if (n->pre) n = n->pre;

}

inline void deleteNodes(Node *&n)

{

	while (n)

	{

		Node *p = n->post;

		while (n)

		{

			Node *d = n->down;

			delete n; n = NULL;

			n = d;

		}

		n = p;

	}

}

int main()

{

	Node *head = new Node; //Dummy head

	while (true)

	{

		Node *it = new Node;

		it->a = getchar();

		if (it->a == '#') break;

		it->b = getchar();

		getchar();

		head->post = it;//initialize chain

		it->pre = head;

		for (int i = 1; i < 52; i++)

		{

			Node *p = new Node;

			p->a = getchar();

			p->b = getchar();

			getchar();

			it->post = p;

			p->pre = it;

			it = p;

		}

		bool checkMove = true;

		while (checkMove)

		{

			checkMove = false;

			it = head->post;

			while (it)

			{

				Node *post = it->post;

				Node *p = it;

				pre3(p);

				if (p && p != head && checkMovable(p, it))

				{

					checkMove = true;

					takeoutNode(it);

					inStackNode(p, it);//调用參数不要乱了

					break;

				}

				p = it;

				pre1(p);

				if (p && p != head && checkMovable(p, it))

				{

					checkMove = true;

					takeoutNode(it);

					inStackNode(p, it);

					break;//题意，理解

				}

				it = post;

			}//while (it)

		}//while (checkMove && piles > 1)

		it = head->post;

		int piles = 0;

		while (it)

		{

			piles++;

			it = it->post;

		}

		if (piles == 1) printf("%d pile remaining:", piles);

		else printf("%d piles remaining:", piles);

		it = head->post;

		while (it)

		{

			printf(" %d", it->size);

			it = it->post;

		}

		putchar('\n');

		deleteNodes(head->post);

	}//while (true)

	delete head;

	return 0;

}