ACM学习历程——UVA127 "Accordian" Patience(栈, 链表)
Description
``Accordian'' Patience |
You are to simulate the playing of games of ``Accordian'' patience, the rules for which are as follows:
Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate neighbour on the left, or matches the third card to the left, it may be moved onto that card. Cards match if they are of the same suit or same rank. After making a move, look to see if it has made additional moves possible. Only the top card of each pile may be moved at any given time. Gaps between piles should be closed up as soon as they appear by moving all piles on the right of the gap one position to the left. Deal out the whole pack, combining cards towards the left whenever possible. The game is won if the pack is reduced to a single pile.
Situations can arise where more than one play is possible. Where two cards may be moved, you should adopt the strategy of always moving the leftmost card possible. Where a card may be moved either one position to the left or three positions to the left, move it three positions.
Input
Input data to the program specifies the order in which cards are dealt from the pack. The input contains pairs of lines, each line containing 26 cards separated by single space characters. The final line of the input file contains a # as its first character. Cards are represented as a two character code. The first character is the face-value (A=Ace, 2-9, T=10, J=Jack, Q=Queen, K=King) and the second character is the suit (C=Clubs, D=Diamonds, H=Hearts, S=Spades).
Output
One line of output must be produced for each pair of lines (that between them describe a pack of 52 cards) in the input. Each line of output shows the number of cards in each of the piles remaining after playing ``Accordian patience'' with the pack of cards as described by the corresponding pairs of input lines.
Sample Input
QD AD 8H 5S 3H 5H TC 4D JH KS 6H 8S JS AC AS 8D 2H QS TS 3S AH 4H TH TD 3C 6S
8C 7D 4C 4S 7S 9H 7C 5D 2S KD 2D QH JD 6D 9D JC 2C KH 3D QC 6C 9S KC 7H 9C 5C
AC 2C 3C 4C 5C 6C 7C 8C 9C TC JC QC KC AD 2D 3D 4D 5D 6D 7D 8D TD 9D JD QD KD
AH 2H 3H 4H 5H 6H 7H 8H 9H KH 6S QH TH AS 2S 3S 4S 5S JH 7S 8S 9S TS JS QS KS
#
Sample Output
6 piles remaining: 40 8 1 1 1 1
1 pile remaining: 52 根据题目意思,两种操作,固然可以看出每个牌堆需要靠栈来实现,故结构体来存放,内部有一个数组成员,可以开52。
然而,对于牌堆来说,似乎要支持访问后一个和前一个牌堆两种操作,故可以使用顺序表,然后访问时忽略0牌堆。
也可以使用双向链表。
下面贴两种方式的代码: 双向链表:
用时432MS:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <string>
#define inf 0x3fffffff
#define eps 1e-10 using namespace std; struct node
{
char Stack[52][2];
int top;
node *pre;
node *next;
}; node *head;
int num; void Del(node *p)
{
node *t;
t = p->pre;
t->next = p->next;
if (p->next != NULL)
p->next->pre = t;
free(p);
} bool Input()
{
num = 0;
head = (node *)malloc(sizeof(node));
node *p = head, *t;
char ch;
ch = getchar();
if (ch == '#')
return 0;
p->Stack[0][0] = ch;
ch = getchar();
p->Stack[0][1] = ch;
p->top = 1;
p->pre = NULL;
p->next = NULL;
num++;
getchar();
for (int i = 0; i < 51; ++i)
{
p->next = (node *)malloc(sizeof(node));
t = p;
p = p->next;
ch = getchar();
p->Stack[0][0] = ch;
ch = getchar();
p->Stack[0][1] = ch;
p->top = 1;
p->next = NULL;
p->pre = t;
num++;
getchar();
}
return 1;
} void Output()
{
if (num == 1)
{
printf("1 pile remaining: %d\n", head->top);
return;
}
printf("%d piles remaining:", num);
node *p = head;
for (;;)
{
printf(" %d", p->top);
if (p->next == NULL)
break;
p = p->next;
}
printf("\n");
} bool Do()
{
node *p, *t;
p = head;
for (;;)
{
t = p;
for (int i = 0; t != NULL && i < 3; ++i)
t = t->pre;
if (t != NULL &&
(t->Stack[t->top-1][0] == p->Stack[p->top-1][0] ||
t->Stack[t->top-1][1] == p->Stack[p->top-1][1]))
{
t->Stack[t->top][0] = p->Stack[p->top-1][0];
t->Stack[t->top][1] = p->Stack[p->top-1][1];
t->top++;
p->top--;
if (p->top == 0)
{
Del(p);
num--;
}
return 1;
}
t = p->pre;
if (t != NULL &&
(t->Stack[t->top-1][0] == p->Stack[p->top-1][0] ||
t->Stack[t->top-1][1] == p->Stack[p->top-1][1]))
{
t->Stack[t->top][0] = p->Stack[p->top-1][0];
t->Stack[t->top][1] = p->Stack[p->top-1][1];
t->top++;
p->top--;
if (p->top == 0)
{
Del(p);
num--;
}
return 1;
}
if (p->next == NULL)
break;
p = p->next;
}
return 0;
} int main()
{
//freopen("test.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
while (Input())
{
while (Do());
Output();
}
return 0;
}
顺序表:
用时:879MS
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#define inf 0x3fffffff
#define esp 1e-10
#define N 100005 using namespace std; struct node
{
int top;
char card[53][3];
}s[52]; bool Input()
{
scanf ("%s", s[0].card[0]);
if (s[0].card[0][0] == '#')
return 0;
s[0].top = 1;
for (int i = 1; i < 52; ++i)
{
s[i].top = 1;
scanf ("%s", s[i].card[0]);
}
return 1;
} void Output()
{
queue <int> q;
for (int i = 0; i < 52; ++i)
if (s[i].top != 0)
q.push(i);
int sum = q.size();
if (sum == 1)
printf ("1 pile remaining: %d\n", s[q.front()].top);
else
{
printf ("%d piles remaining:", sum);
int k;
while (!q.empty())
{
k = q.front();
q.pop();
printf (" %d", s[k].top);
}
printf ("\n");
}
} void qt()
{
int p = 0;
for (;;)
{
if (p == 52)
break;
if (s[p].top != 0)
{
int one = -1, two = -1;
int flag = 0, j = p-1;
for (;;)
{
if (j < 0)
break;
if (s[j].top != 0)
{
flag++;
if (flag == 3)
{
one = j;
break;
}
if (flag == 1)
two = j;
}
j--;
}
if (one != -1)
{
if (s[p].card[s[p].top-1][0] == s[one].card[s[one].top-1][0] ||
s[p].card[s[p].top-1][1] == s[one].card[s[one].top-1][1])
{
strcpy (s[one].card[s[one].top], s[p].card[s[p].top-1]);
s[p].top--;
s[one].top++;
p = 0;
continue;
}
}
if (two != -1)
{
if (s[p].card[s[p].top-1][0] == s[two].card[s[two].top-1][0] ||
s[p].card[s[p].top-1][1] == s[two].card[s[two].top-1][1])
{
strcpy (s[two].card[s[two].top], s[p].card[s[p].top-1]);
s[p].top--;
s[two].top++;
p = 0;
continue;
}
}
}
++p;
}
} int main()
{
//freopen ("test.txt", "r", stdin);
while (Input())
{
qt();
Output();
}
return 0;
}
ACM学习历程——UVA127 "Accordian" Patience(栈, 链表)的更多相关文章
- ACM学习历程——UVA11234 Expressions(栈,队列,树的遍历,后序遍历,bfs)
Description Problem E: Expressions2007/2008 ACM International Collegiate Programming Contest Unive ...
- [刷题]算法竞赛入门经典(第2版) 6-9/UVa127 - "Accordian" Patience
题意:52张牌排一行,一旦出现任何一张牌与它左边的第一张或第三张"匹配",即花色或点数相同,则须立即将其移动到那张牌上面,将其覆盖.能执行以上移动的只有压在最上面的牌.直到最后没有 ...
- ACM学习历程——UVA 127 "Accordian" Patience(栈;模拟)
Description ``Accordian'' Patience You are to simulate the playing of games of ``Accordian'' patie ...
- ACM学习历程——UVA11111 Generalized Matrioshkas(栈)
Description Problem B - Generalized Matrioshkas Problem B - Generalized Matrioshkas Vladimir wo ...
- ACM学习历程——UVA442 Matrix Chain Multiplication(栈)
Description Matrix Chain Multiplication Matrix Chain Multiplication Suppose you have to evaluate ...
- ACM学习历程——ZOJ 3829 Known Notation (2014牡丹江区域赛K题)(策略,栈)
Description Do you know reverse Polish notation (RPN)? It is a known notation in the area of mathema ...
- 完成了C++作业,本博客现在开始全面记录acm学习历程,真正的acm之路,现在开始
以下以目前遇到题目开始记录,按发布时间排序 ACM之递推递归 ACM之数学题 拓扑排序 ACM之最短路径做题笔记与记录 STL学习笔记不(定期更新) 八皇后问题解题报告
- ACM学习历程—HDU 5512 Pagodas(数学)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5512 学习菊苣的博客,只粘链接,不粘题目描述了. 题目大意就是给了初始的集合{a, b},然后取集合里 ...
- ACM学习历程—HDU5521 Meeting(图论)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5521 学习菊苣的博客,只粘链接,不粘题目描述了. 题目大意就是一个人从1开始走,一个人从n开始走.让最 ...
随机推荐
- uboot之bootm以及go命令的实现
本文档简单介绍了uboot中用于引导内核的命令bootm的实现,同时分析了uImage文件的格式,还简单看了一下uboot下go命令的实现 作者: 彭东林 邮箱: pengdonglin137@163 ...
- 用array_search 数组中查找是否存在这个 值
#判读里面是否还有id=1的超级管理员 $key=array_search(1, $ids); #判读这个是否存在 if($key!==FALSE){ #如果存在就unset掉这个 unset($id ...
- SQL ROW_NUMBER() 分页使用示例
ALTER PROC [dbo].[TestProPage] , AS BEGIN SELECT * FROM (SELECT *,ROW_NUMBER() OVER(ORDER BY IndexID ...
- 用变量a给出下面的定义。[中国台湾某著名CPU生产公司2005年面试题]
(1)一个整型数(An integer)(2)一个指向整型数的指针(A pointer to an integer)(3)一个指向指针的指针,它指向的指针是指向一个整型数(A pointer to a ...
- Grunt是什么,以及它的一些使用方法
♥什么是Grunt Grunt和 Grunt 插件是通过 npm 安装并管理的,npm是 Node.js 的包管理器.grunt是基于node 更多插件请访问:http://www.gruntjs.n ...
- android菜鸟学习笔记15----Android Junit测试
Android中的Junit测试与Java Junit测试有所不同,不能简单的使用标注…… 假设写了一个MathUtils类,有两个静态方法: public class MathUtils { pub ...
- 关于js全局变量数组push数据时dom中无数据的问题
今天着实悲催,这问题整了好几个小时才解决.废话不多说,上问题. 一开始我定义了许多全局变量放在me下. var me = { dgOrderDetails: null, dgVisitNumbers: ...
- Mac下终端常用命令
一.删除文件 1 打开终端应用程序 2 输入命令:sudo (空格) rm (空格)-r (空格)-f (空格)(注意-f后面还有空格),还要注意,全部小写. 3 把你要删的文件或者文件夹用mouse ...
- Linux就该这么学--命令集合11(配置系统相关信息)
1.配置主机名称: 查看主机名: hostname 修改主机名: vim /etc/hostname 2.配置网卡信息: 在红帽RHEL6系统中网卡配置文件的前缀为“ifcfg-eth”,第一块即为“ ...
- ABAP-创建信息记录
CALL FUNCTION 'ME_INITIALIZE_INFORECORD'. CALL FUNCTION 'ME_DIRECT_INPUT_INFORECORD' *&--------- ...