• UVa 127


    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('
    ');
    		deleteNodes(head->post);
    	}//while (true)
    	delete head;
    	return 0;
    }



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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/5417415.html
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