UvaOJ 127 - "Accordian" Patience

题目

题意规则挺多的，但是读懂了还是清晰的，这里就不写了。

模拟题。我开了个双向链表，用struct 的 oop写的，没access control 看着挺乱但是写着还是挺爽的。130行...1Y。

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

#define MAXN 53  //52cards
struct Card
{
	Card() {}
    Card(char r, char s) { rank=r; suit = s; }
    char rank;
    char suit;
	int isSame(Card c)
	{
		return rank==c.rank || suit==c.suit;
	}
};
struct Pile      //node
{
    Pile() { top = 0; left = right = NULL; }
    int top;
    Card cards[MAXN];
    Pile *left, *right;
	int isSame(Pile* p)
	{
		return p->cards[p->top-1].isSame(cards[top-1]);
	}
	void add(Card c)
	{
		cards[top++] = c;
	}
	Card pop()
	{
		return cards[--top];
	}
	int isEmpty()
	{
		return top == 0;
	}
};
struct List
{
    List() 
    { 
        size = 0; head = tail = NULL; 
    }
    int read()		//初始化读入
    {
		size = 52;
		head = tail = NULL; 
        char buf[3];
		Pile* tp = NULL;
        for(int i=0; i<52; i++)
        {
            scanf("%s", buf);
			if(buf[0] == '#') return 0;		//end of input
            Pile* p= new Pile();
			p->add(Card(buf[0], buf[1]));
            if(!i) head = p;
			else if(i==51) tail = p;
			if(tp) 
			{
				tp->right = p;
				p->left = tp;
			}
			tp = p;
        }
		return 1;
    }
    Pile *head, *tail;       //head->->tail
    int size;

	void del(Pile* p)
	{
		if(p->left)
		{
			p->left->right = p->right;
		}
		if(p->right)
		{
			p->right->left = p->left;
		}
		size--;
		delete p;
	}
	void move(Pile* p, Pile* to)	//p move to 
	{
		to->add(p->pop());
		if(p->isEmpty())
			del(p);
	}
	int solve()	//return 0 to finish
	{
		Pile* tp = head->right;
		for(; tp; tp = tp->right)
		{
			if(tp->left && tp->left->left && tp->left->left->left && tp->left->left->left->isSame(tp))	//短路
			{
				move(tp, tp->left->left->left);
				return 1;
			}
			if(tp->left && tp->left->isSame(tp))
			{
				move(tp, tp->left);
				return 1;
			}
		}
		return 0;
	}
	void print()
	{
		printf("%d pile", size);
		printf(size>1? "s": "");
		printf(" remaining:");
		for(Pile* p=head; p; p=p->right)		//head指针不会变
		{
			printf(" %d", p->top);
		}
		putchar('\n');
	}
}row;

int main()
{
	while(row.read())
	{
		while(row.solve()){}
		row.print();
	}
}